 Phyllis Fleming Physics Physics 107
 Answers - Angular Motion
 1 a. For constant linear velocity, the ratio of the distance x moved to the time t,  x/t = constant = v  or  x = vt.  b. For constant angular velocity, the ratio of the angle Θ moved to the time t, Θ/t = constant = ω   or   Θ= ωt.
 2 a. Constant linear acceleration a equals the change in linear velocity      divided by the time to change the velocity, or a = (v - vo)/(t - 0)  or   v = vo + at.  b. Constant angular acceleration α equals the change in angular velocity      divided by the time to change the velocity, or α = (ω - ωo)/(t - 0)  or   ω = ωo + αt.

3.

 a. dx/dt = v = vo + at  or b. dΘ/dt = ω = ωo + αt  or  4 ω(t) = ωo + αt. ω(4.0 s) = πs-1 + (4 π s-2)(4.0 s) =17 πs-1. Θ(t) = Θo + ωot + 1/2 αt2. Θ(4.0s) = 0 + ( π s-1)(4.0s) + 1/2 (4 πs-2)(16s2) = 36 π. In one rotation, the wheel turns through an angle of 2 π radians. The number of turns made by the wheel = 36 π/2 π = 18.
 5 ω(t) = dΘ/dt = d(b + ct + et2)/dt = c + 2et. ω(t1) = c + 2et1 α(t) = dω/dt = d(c + 2et)/dt= 2e = constant = α(t1). v(t1) = ω(t1)r = (c + 2et1)r. Tangential acceleration, a(t1) = α(t1)r = 2er. Centripetal or radial acceleration(t1) = { v(t1)}2/r = (c + 2et1)2r2/r = (c + 2et1)2r.
 6 If v = ω x r,  then v is perpendicular to  ω  and v is perpendicular to r, or v is perpendicular to the plane that contains  ω  and r.  In the figure above you see that  ω  is out of the page and v is in the xy-plane, so v is perpendicular to ω.  Since v is always tangent to the path, it is also perpendicular to r. The plane of  ω  and r is the xz-plane, so v is perpendicular to the plane that contains  ω  and r.  If you point the fingers of your right hand in the direction of  ω  (out of the page) and curl them in the direction of r, your outstretched thumb points in the direction of v.  Also, v = ωr sin ω,r = ωr sin 90o = ωr.  ω x r gives both the correct direction and magnitude of v. a = ω x (ω x r) = ω x v from above.  a is into the center of the circle and perpendicular to both  ω  and v.  If you point the fingers of your right hand in the direction of  ω  (out of the page) and curl them in the direction of v, your outstretched thumb points in the direction of a or into the center of the circle.  a = ωv sin w,  v = ωv sin 90o =  ωv = (v/r)(v) = v2/r = ω2r.  The direction and magnitude of the centripetal acceleration is given by: a  =  ω  x  (ω x r) =  ω  x  v.
 7 The constant angular velocity  ω = (Θ - 0)/(t - 0) = Θ/t.  In time t,  both points 1 and 2 rotate through angle Θ so both points have the same angular velocity. For point 1,  v1 = s1/t  and for point 2,  v2 = s2/t.  Since s2 > s1,  v2 > v1.  This is also shown from v1 = ωr1 and v2 = ωr2.  Again since r2 > r1,  v2 > v1.
 8 For one particle of mass m and velocity v,  the kinetic energy K = 1/2 mv2. For all of the particles that make up the disk from i = 1 to i = N, K = 1/2 m1v12 + 1/2 m2v22 + 1/2 m3v32 + . . . .1/2 mNvN2. where I is the moment of inertia of the rigid body.
 9 The area of the plate  A = ab  (Fig. 3 above). The mass per unit area  σ = M/A = M/ab. The differential area with height b and thickness dx is dA = b dx. The mass of this area is the mass per unit area times dA. That is dm = σ dA = (M/ab) (b dx) = M dx/a. Its moment of inertia  dI = x2 dm = x2 (M dx/a) = (M/a) x2 dx. The moment of inertia for the entire plate about the Y-axis is: 10 The differential volume of the disk of radius x and thickness dy is dV = πx2 dy. Its mass dm = the mass per unit volume times the volume =                dm = ρ dV = {M/(4 πR3/3)} { πx2 dy).      dI =1/2 dm x2 =                               1/2{M/(4 πR3/3)}{ πx2 dy)x2. From Fig. 4, we see that  x2 = (R2 - y2)  and  dI = 3M/8R3 (R2 - y2)2 dy. 11 I = iσ miri2.   The moment of inertia of the plate about the Y-axis is greater than the moment of inertia about CC’ because the mass of the plate is less concentrated about the Y-axis than CC’.  In other words, there are greater values of r for the Y-axis calculation. Again  dI = (M/a) x2 dx.  But taking x = 0 at C’, we must integrate from -a/2 to a/2: CC’ passes through the center of mass of the plate. The parallel axis theorem states that the moment of inertia about a parallel axis a distance d from an axis through the center of mass equals the moment of inertia about the center of mass + Md2.  In this case  IY = ICC’ + M(a/2)2 since the distance between the Y-axis and CC’ is  a/2. 1/3 Ma2 = ICC’ + Ma2/4   or λCC’ = Ma2(1/3 - 1/4) = 1/12 Ma2.
 12  13 The string is always tangent to the surface of the cylinder. The angle between the tension and the radial vector is 90o.  The torque applied to the cylinder is constant and has magnitude τ = RT sin 90o = 0.040 m(3.0 N))(1) = 0.12 N-m. The moment of inertia of the cylinder about its axis is I = 1/2 MR2 = 1/2(30 kg)(0.040 m)2 = 2.4 x 10-2 kg-m2. α = τ/I = 0.12 N-m/2.4 x 10-2 kg-m2 = 5.0 s-2. The angular speed at  t = 2.0 s  is ω = ωo + αt = 0 +(5.0 s-2)(2.0 s) = 10 s-1.
 14 Applying Newton's second law to the hanging block,    Fnet = ma mg - T = ma              (Equation 1) For the accelerated disk and taking clockwise rotation as positive, τnet = Iα RT sin 90o = 1/2 MR2 α = 1/2 MR2 (a/R) T = 1/2 Ma          (Equation 2) Substituting Eq. 2 into Eq. 1: mg - 1/2Ma = ma a = mg/(M/2 + m)=(0.50 kg)(10 m/s2)/(1.0/2 + 0.5)kg = 5.0 m/s2. Applying Newton's second law to the disk, (Fnet)y = may N - Mg - T = m(0) = 0  or N = Mg + T        (Equation 3) From Eq. 2, T = 1/2 Ma = 1/2 (1.0 kg)(5.0 m/s2) = 2.5 N N = Mg + T = (1.0 kg)(10 m/s2) + 2.5 N = 12.5 N
 15   16 As illustrated in Fig. 8 above,    τabout pivot  =  Iabout pivot α L/2(Mg)sin 90o = 1/3 ML2 α                      α = 3/2 (g/L) a = αL = 3/2(g/L) (L) = 3g/2
 17 The moment of inertia of the disk about the pivot in Fig. 9 above is I = ICM + Md2, where d the distance of the center of mass from the pivot is R.  Thus, Ipivot = 1/2 MR2 + MR2 = 3MR2/2. For the hoop in part (c) Ipivot = MR2 + MR2 = 2MR2. Take the "final" gravitational potential energy Uf = 0.  Then the initial gravitational potential energy, when it is in the position shown by the solid circle,  Ui = MgR,  since the center of mass of the disk in its initial position is a distance R above its "final" position.  Because it is initially at rest,  Ki = 0. From conservation of energy,  Ui        +          Ki          =         Uf         +         Kf MgR      +          0          =          0         +   1/2(3/2MR2)ω2 ω = 2(g/3R)1/2 and at the center of mass  v = ωR = 2(gR/3)1/2 At the bottom of the disk,  v = ω2R = 4(gR/3)1/2 Now, MgR        +        0          =           0          +   1/2(2MR2)ω2     and ω = (g/R)1/2 At center of mass v = R(g/R)1/2 = (gR)1/2. At bottom v = 2(gR)1/2.
 18 The center of mass of the disk translates and the disk rotates. The total kinetic energy of the disk is the sum of its translational and rotational kinetic energy = 1/2 Mv2 + 1/2 Iω2 = 1/2 Mv2 + 1/2(1/2 MR2)(v/R)2 = 3/4 Mv2.
 19 The forces acting on the spool are F, its weight W, normal force N, and the frictional force f.  f acts to the right because when you pull it to the right, the spool would tend to slip backwards. A frictional force then acts on it to the right. Taking the axis of rotation about the center of the sphere means that the weight and normal force produce no torque since their lines of action passes through the center of the sphere. The force F produces clockwise rotation, while f produces counterclockwise rotation. τ = r x F For translation, Fnet = Ma F + f = Ma            (Equation 1) For rotation, τnet = Iα RF -Rf = 1/2 MR2 (a/R)   or F - f = 1/2 Ma                       (Equation 2) Adding Eq. 1 and Eq. 2: 2F = 3/2 Ma   or   a = 4F/3M From Eq. 1, F + f = Ma,  with Ma = 4F/3,   F + f = (4F/3)   or   f = F/3.
 20 Use conservation of energy. This time take the potential energy of the initial state (Fig. 11i) as the zero of potential. The final potential energy state (Fig. 11f) will be negative. The leg of the H about which the system rotates remains at rest. Its potential energy is zero at all times. The center leg of the H center of mass moves down to L/2 and the other leg moves down L. The kinetic energy is gained by the center leg with moment of inertia 1/3 ML2 about one end and the leg on the right whose mass acts at the center of mass and has the moment of inertia of a point particle ML2. Ui + Ki =                Uf          +                          Kf  0 + 0  =  - (MgL/2 + MgL)  +  1/2(1/3 ML2)ω2 + 1/2(ML2)ω2 MgL(1/2 + 1) = (1/2 ML2ω2)(1/3 + 1)   3/2(g/L)1/2 = ω
 21 The frictional force f acts at a point, not through a distance, so it does no work and energy is conserved. We take the gravitational potential energy of the sphere equal to 0 at the bottom of the hill, that is,  Uf = 0. The center of mass of the sphere at the top of the hill is a distance  h above where it is at the bottom of the hill: Ui = Mgh. As the sphere rolls down the hill, it has kinetic energy of translation of the center of mass and it has rotational kinetic energy. From conservation of energy, Ui + Ki = Uf + Kf Mgh + 0 = 0 + 1/2 Mv2 + 1/2 Iω2 = 0 + 1/2 Mv2 + 1/2(2/5MR2)(v/R)2 Mgh = 7/10 Mv2   or   v = (10gh/7)1/2      (Fnet)x = ma Mg sin Θ - f = Ma        (Equation 1) Take the torques about the center of mass. Since the weight and normal force pass through this axis, they produces no torque.               τ = Ia Rf sin 90o = (2/5 MR2)(a/R)               f = 2/5 Ma                                (Equation 2) Substituting Eq. 2 into Eq. 1: Mg sin Θ - 2/5 Ma = Ma  or  a = 5g sin Θ/7 From Eq. 2, f = 2/5 Ma = (2/5)M(5g sin Θ/7) = 2 Mg sin Θ/7 v2 = vo2 + 2as = 0 + 2(5g sin Θ/7)s           = (10g/7)(s sin Θ)           = (10g/7)(h)       v = (10gh/7)1/2, as found in Part (a).
 22 Assume the center of mass of the pencil is at its center L/2 (as shown in Fig for #22 above).  From conservation of energy,     Ui    + Ki = Uf +    Kf mgL/2 +  0 =  0 + 1/2 Iω2 ω = (mgL/I)1/2 = (mgh/1/3 mL2)1/2 = (3g/L)1/2 vtop of pencil = ωL = (3g/L)1/2L = (3gL)1/2
 23 From Newton's second law: Fnet = ma F - f = Ma                           (Equation 1) For rotation about center,      τnet = Ia Rf sin Θ = (1/2 MR2)(a/r)           f = 1/2 Ma                       (Equation 2) Adding Eq. 2 to Eq.1: F = 3/2 Ma   or   a = 2F/3M         (Equation 3) Substituting Eq. 3 into Eq. 2: f = 1/2 M (2F/3M) = F/3

24. 1.  Fnet = ma f = Ma            (Equation 1) τnet = lα - Rf sin 90o = Iα  for clockwise rotation                              (Equation 3) a = dv/dt = f/M α = dω/dt = -fR/I = -5f/2MR (Equation 2) (Equation 4)

Substituting Eq. 2 into Eq. 4:
ωT - ωo = - 5vT/2R
At  t = T, vT = ωTR   and
ωT - ωo = -5ωT/2  or
ωT = 2ωo/7 = 2(70 rad/s)/7 = 20 rad/s.
2. Ko = 1/2 Iωo2 = 1/2 (2/5 MR2o2                 (Equation 5)
= 1/10(2.0 kg)(0.10 m)(70 rad/s)2 = 98J

KT = 1/2 MvT2 + 1/2 Iω2                        (Equation 6)
= 1/2 M(RωT)2 + 1/2(2/5 MR2)(ωT)2

= 7/10 MR2ωT2
KT = 7/10 MR2(2ωo/7)2 = 1/7 Iωo2 = 2/7 Ko = 2/7(98 J) = 28 J.

3. For 0 < t < T,  the sphere slips a distance ds = (Rω - v)dt.  To justify this we see that at  t = 0,  v = 0 and ds = Rωo.  At  t = T,  ω  =  ωand  v = RωT,   so ds = 0. Work done by friction = dW = f ds = fRωdt - fv dt.  For the first f substitute from Equation 3,  f = - Iα/R  and for the second  f  from Equation 1,  f = Ma.

dW = - Iαωdt - Mavdt = - I(dω/dt)ωdt - M(dv/dt)vdt = - (Iω + Mv)dt. From Equation 5 and Equation 6,   W = Ko -  KT.

 25 Use conservation of energy to solve the problem. Take the gravitational potential energy equal to 0 at the bottom of the loop-the-loop. Remember that the sphere will have both rotational and translational kinetic energy. The sphere is released at the initial position i so the kinetic energy there is zero. The sphere will make it around the loop-the-loop if it can stay on the loop as it moves down from the top of the loop. For this reason we take the top of the loop as the final position f. Ui + Ki = Uf + Kf mgh + 0 = 2mgR + 1/2 mv2 + 1/2 Iω2   or mgh = 2mgR + 1/2 mvf2 + 1/2 (2/5mR2)(vf/R)2 mgh= 2mgR + 7/10 mvf2                            (Equation 1) At the top, the acceleration is in toward the center.     Fnet = ma mg + N = mvf2/R. For the minimum height h, we want the minimum velocity vf,  so we set the normal force N = 0.  Then, mg = mvf2/R  or  mvf2 = mgR         (Equation 2) Substituting Eq. 2 into Eq. 1: mgh = 27/10 mgR  or  h = 2.7R. At P, the potential energy = mgR. Now, Ui + Ki = UP + KP mg(2.7R) + 0 = mgR + 0.7mvP2 mg(2.7 - 1.0)R = 1.7mgR = 0.7mvP2  or  mvP2 = 17/7 mgR. At P, the normal force produces the centripetal acceleration into the center of the circle.  At P, mg is down. Fnet = ma N = mvP2/R = {(17/7)mgR}/R = 17/7 mg
 26 L = r x mv.  L is out of the page (Fig. for #26 above). L = rmv sin r,  v= rmv sin 90o = rmv. Since v = ωr,  L = (mr2)ω = Iω.
 27  28 L = r x mv.  L is into the page (Fig. 13 above). L = rmv sin r,  v = mv(r sin r, v) = mvb.
 29 L = r x mv. L is out of the page for both particles (Fig. 14 above). L = rmv sin r,  v = rmv sin 90o. Total L = 0.50 m(5.0 m/s)(3.0 + 4.0) kg = 17.5 kg-m2/s.
 30 L = r x mv dL/dt = d(r x mv)/dt = r x d(mv)/dt + dr/dt x mv          = r x F + (v x mv)          = r x F + (0)          = r x F. Note: d(mv)/dt = dp/dt = F and (v x v) = 0.
 31 τ = R x mg is toward the right (Fig. 15 above) τ= Rmg sin 90o = Rmg L = 1/2 MR2 ω + Rmv = 1/2 MR2(v/R) + Rmv = vR(M/2 + m) dL/dt = d[vR(M/2 + m)]/dt = R(M/2 + m) dv/dt dL/dt = R(M/2 + m)a = Rmg = τ. a = mg/(M/2 + m)
 32 At the origin, r = 0 and L = r x mv = 0. At highest point,  vy2 = 0 = (vo sin Θ)2 -2g(ymax - yo). Since yo = 0,  ymax = (vo sin Θ)2/2g. rb = rbx i + (vo sin Θ)2/2g j. vb = vo cos Θ = vox because vby = 0 and there is no acceleration in the X-direction. vb =vo cos Θi. Lb = rb x mvb = (rbx i + (vo sin Θ)2/2g j) x m(vo cos Θi)      = (rbxmvo cos Θ)(i x i) + (mvo3 sin2 Θ cos Θ/2g) (j x i)      = (rbxmvo cos Θ)(0) + (mvo3 sin2 Θ cos Θ/2g)(-k)      = - (mvo3 sin2 Θ cos Θ/2g)k The length of rc  is the range R = 2vo2 sin Θcos Θ/g. rc = 2vo2 sin Θcos Θ/g i. vc = vo(cos Θi - sin Θ j). When the projectile leaves the ground and returns to the ground, the X-component of the velocity remains the same (constant horizontal velocity) and the Y-component has the same magnitude as the initial Y- component of velocity, but now it is negative. Lc = rc x mvc     = (2vo2 sin Θcos Θ/g) i x mvo (cos Θi - sin Θj)     = - (2mvo3 sin2 Θcos Θ/g)k. ΔL is in -k direction. τ = r x -mgj = (rxi + ryj) x -mgj = -rxmg k.

33. 1. For a point particle, L = (r x mv). For the initial angular momentum, the magnitude of Li = rmv sin r, v = mv (r sin r, v) = mvR. After the collision, the particle sticks to the edge of the disk and moves with the disk with angular velocity ω. The moment of inertia of the disk is 1/2 MR2 and that of the particle is mR2. Since angular momentum is conserved,

 Li = Lf mvR = 1/2 MR2 ω + mR2ω = R2ω(1/2 M + m) = (R2ω/2)(M + 2m) ω = 2mv/R(M + 2m).

2. Ki = 1/2 mv2.
Kf = 1/2 ω2 (Idisk + Iparticle) = 1/2 ω2(1/2 MR2 + mR2)
= 1/2 {4m2v2/(M + 2m)2R2}(M + 2m)R2/2 = m2v2/(M + 2m).

Energy lost = Ki - Kf = 1/2 mv2 - m2v2/(M + 2m) = mMv2/2(M + 2m).

 34 Take the axis at the force center (Fig. 17 above). The repulsive force acts along the line connecting the two masses, passing through the axis, so the net torque equals zero and angular momentum is conserved. Li = rimvi sin ri, vi = (ri sin ri, vi)mvi = bmvi.    Lf = dminmvf. From conservation of momentum, bmvi = dminmvf and vf = bvi/dmin     (Equation 1) From conservation of energy,           Ui + Ki = Uf + Kf 0 + 1/2 mvi2 = A/dmin + 1/2 mvf2         (Equation 2) Substituting Eq. 1 into Eq. 2: 1/2 mvi2 = A/dmin + 1/2 m(bvi/dmin)2 dmin2 - (2A/mvo2)dmin - b2 = 0 dmin = {(2A/mvo2) ± [(2A/mvo2)2 + 4b2]1/2}/2 dmin = (A/mvo2) + [(A/mvo2)2 + b2]1/2
 35 Curl the fingers of your right hand clockwise to find the angular momentum L to the right (Fig. 18 above).  τ = r x F is out of the page. The torque due to the weight of the wheel is into the page, but we have made the torque due to F greater and the resultant torque is out of the page.  Since τ = ΔL/ Δt,  ΔL will be in the direction of τ, out of the page or to the right as you view it from above. The wheel turns to the right.
 36 d(L2)/dt = d(L . L)/dt = L. dL/dt + dL/dt. L = 2L . dL/dt = 2L . τ  = 2Lτ cos L, τ.   If the angle between L and τ is 90o,  d(L2)/dt = 0 and the magnitude of the angular momentum remains constant.
 37 Taking the axis at A,  F produces no torque because it passes through the axis. τ = r x mg. The torque due to mg is into the page, as shown in Fig. for #37(a), a side view, and to the left in Fig. for #37(b), a top view. The magnitude of the torque  τ = rmg.  Since τ = dL/dt or dL = τ dt, dL is to the left as shown in Fig. for #37(c), a top view. The angle turned through by L is dΦ = τ dt/L or dΦ/dt = mgr/L = mgr/Iω. The wheel precesses counterclockwise with dΦ/dt so the platform must rotate with dΦ/dt = mgr/Iω  counterclockwise for the wheel to maintain a fixed position relative to the platform.

38. For equilibrium, Στ = 0.
Taking the axis at the bottom of the ladder, with τ = r F sin r, F,

Στ = L Fwall sin Θ - (L/2)W cos Θ = 0  or
tan Θ= W/2 Fwall                              (Equation 1)
Note that neither N nor f contribute to the torque when the axis is at the bottom of the ladder because both forces pass through the axis. Also for equilibrium,

 σ Fx = 0       and σ Fy = 0 f - Fwall = 0 N - W = 0 f = Fwall N = W but f = µN so f = µW and µW = Fwall     (Equation 2)

Substituting Eq. (2) into Eq. (1):
tan Θ= W/2µW = 1/2µ = 1/0.80 = 1.25.   Θ= 51.3o.

 39 Take the axis at O in the above figure. The torque due to Mg = rMg sin Θ = Mg (r sin Θ) = MgR clockwise. The torque due to the normal force N = rN sin Θ= NR counterclockwise. Since there is no acceleration in the vertical direction,  N - Mg = 0  or  N = Mg. The torque due to Mg equals the torque due to N so that the net torque due to these two forces is zero. The torque due to the frictional force = fx sin f, x-axis = fx sin 0o = 0. Since no net torque acts on the sphere, angular momentum is conserved. The initial angular momentum LI = Iωo = 2/5 MR2ωo. The final angular momentum when it is both rotating and the center of mass moves with velocity v = Lf = Iω + MRv. For the sphere to rotate without slipping, v = ωR or Lf = 2/5 MR2ω + MR2ω. From conservation of angular momentum,               LI = Lf 2/5 MR2ωo = 2/5 MR2ω + MR2ω = 7/5 MR2ω         2ωo/ 7 = ω
 40 Li = rmvi sin r,  vi = (r sin r, vi)mvi =(L/2)mvi. Just after the putty sticks to the point particle on the right: Lf = (2Ifor M’s + Ifor m) ω = [2M + m](L/2)2ω.           Li = Lf mvi(L/2) = [2M + m](L/2)2ω. ω = 2mvi/(2M + m)L    = 2(0.05 kg)(3.00 m/s)/[2(0.975) + 0.05]kg(1.0 m)    = 0.15 s-1. As the rod rotates the total mechanical energy of the system remains constant. The sum of the potential energy and kinetic energy of the system remains constant. Take the zero of potential energy when the rod is horizontal. Thus, Uib = 0. As one of the point objects of mass M rises, the other lowers, keeping the potential energy of these two constant. For this reason, we need only be concerned about the putty. When the putty moves through 90o to reach its lowest point on its path, it gains kinetic energy and loses gravitational potential energy (the potential energy becomes negative). When it swings back up, it loses kinetic energy while gaining potential energy. When the rod with the putty on the end rises through an angle Θ its potential energy increases and its kinetic energy goes to zero. The potential energy at angle Θ is: Ufb = mgh = mgL/2 sin Θ  (Fig. b for #40 above). Uib +             Kib                 =     Ufb          + Kfb  0   + 1/2[2M + m](L/2)2ω2 = mgL/2 sin Θ +  0 [2M + m](L/2mg)ω2 = sin Θ= 0.045.   Θ= 2.6o. Total angle swung through = 180o + 2.6o = 182.6o.

 Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 April 17, 2003