Answers - Angular Motion

1.

2.

3.

         dx/dt = v = v_o + at, or                         dΘ/dt = ω = w_o + at, or

4.

(a)  ω(t) = ω_o + αt.   ω(4.0 s) = π s^-1 + (4 π s^-2)(4.0 s) =17 π s^-1

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.

(a) ω(t) = dΘ/dt = d(b + ct + et²)/dt = c + 2et.  ω(t₁) = c + 2et₁

(b) α(t) = dω/dt = d(c + 2et)/dt= 2e = constant = α(t₁)

(c) v(t₁) = ω(t₁)r = (c + 2et₁)r

(d) tangential acceleration, a(t₁) = α(t₁)r = 2er

(e) centripetal or radial acceleration(t₁) ={ v(t₁)}²/r = (c + 2et₁)²r²/r
     = (c + 2et₁)²r

6.

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.

7.

Angular acceleration α = (ω_10s - ω_o)/(3 s - 0) =(6.0 rad/s - 0)/(3 s) = 2 rad/s². Angle turned through from t = 0 to t = 10 s = ω_ot + 1/2 αt² = 0 +
1/2(2 rad/s²)(100 s²) = 100 rad.  ω(10 s) = ω_o + αt = 0 + (2.0 rad/s²)(10s) =
20 rad/s.  After t = 10 s, α = 0.  Angle turned through from t = 0 to t =
20 s Θ (20 s) = Θ_10s + ω_10s + 1/2 αt² = 100 rad + 20 rad/s(10 s) + 0 = 300 rad.

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 m_Nv_N²
= 1/2 m_iv_i² = 1/2 m_i (ωr_i)² = 1/2 m_iω²r_i² = 1/2 ω² m_ir_i² = 1/2 ω²I
=1/2 Iω², where I is the moment of inertia of the rigid body.

9.

(a) I = mL² + m(4L²) = 5mL².

(b) In general, K = 1/2 Iω².

     For object closest to the axis O,
                        K = 1/2 mL²ω².

     For object farthest from O,
                        K = 1/2 4mL²ω².

               Total K = 1/2 L²ω² (1 + 4) = 5/2(L²ω²).

10.

      τ = r x F
      τ₁ = r₁F₁ sin 30⁰ = 2.0 m(1.5 N)(0.5) = 1.5 N-m

Point the fingers of your right hand in the direction of r₁ and rotate it into F₁ and your thumb points into the page.  τ₁ is into the page.

      τ₂ = r₂F₂ sin 30⁰ = 1.0 m(1.0 N)(0.8) = 0.5 N-m

Point the fingers of your right hand in the direction of r₂ and rotate it into F₂ and your thumb points out of the page.  τ₂ is out of the page.  Resultant torque τ= (1.5 - 0.8)N-m = 0.7 N-m into the page.

11.

(a) Moment of inertia I of a rod about one end is 1/3 mL².
K = 1/2 Iω² = 1/2(1/3mL²)ω².

(b) Let the initial position i be at the lowest position and the final position f be at the highest position. Take the gravitational potential energy of the center of the rod to be zero at the lowest position, that is, U_i = 0. The rod comes momentarily at rest at the highest position, that is, K_f = 0. From conservation of energy, U_f - U_i = (mg)change in height of center of mass = K_f - K_i = 1/6 (mL²)ω². Change in height of center of mass = (L²ω²/6g).

12.

Assume the center of mass of the pencil is at its center L/2 (Fig for # 12).
From conservation of energy,

13.

(a)	See Fig. for #13 to the right,
	τabout pivot = Iabout pivot α
	L/2(Mg)sin 900 = 1/3 ML2 α
	α = 3/2 (g/L)

(b) a = α L = 3/2(g/L) (L) = 3g/2

14.

15.

The forces on the wheel are shown in Fig. for #15a above.  F_h and F_v are the horizontal and vertical forces, respectively, of the curb on the wheel. These forces are not shown in Fig. for #15b above because they do not produce a torque about axis O. Figure for #15b shows the perpendicular distance to F and mg from O. The perpendicular distance from the axis O to the line of action of force F is (R - h).  The torque due to F = (R - h)F into the page.  The perpendicular distance from O to the line of action of mg =
{R² - (R - h)²}^1/2 ={R² - R² + 2Rh - h²}^1/2 = {2Rh - h²}^1/2.  The torque due to mg = {2Rh - h²}^1/2mg out of the page.  To rotate the wheel clockwise over the curb, the torque into the page must be slightly larger than the torque out of the page: (R - h)F > {2Rh - h²}^1/2mg, or F must be slightly greater than
{2Rh - h²}^1/2mg/(R - h).

16.

(a) L = r x mv. Because r x v is out of the page (Fig. for #16), L is out of the      page.  L = rmv sin r,   v= rmv sin 90⁰ = rmv.

(b) Since v = ωr, L = (mr²)ω = Iω.

17.

18.

19.

20.

	τ = r x F
	τ = r F sin r, F

For the satellite orbiting about the earth as its axis, r, F = 180⁰.
sin 180⁰ = 0 and the torque equals zero. When no torque acts on a system, its angular momentum is conserved. The angular momentum of the satellite at P₁ = the angular momentum of the satellite at P_2.

21.

(a) 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, U_f = 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: U_i = 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,

(b)

Take the torques about the center of mass. Since the weight passes through this axis, it produces no torque.

v = (10gh/7)^1/2 as found in Part (a).

22.

For equilibrium, Στ = 0.

Taking the axis at the bottom of the ladder, with τ = r F sin r, F,

                         Στ = L F_wall sin Θ - (L/2)W cos Θ = 0
               or   tan Θ = W/(2 F_wall)                          (Equation 1)

Note that neither N nor f contributes to the torque when the axis is at the bottom of the ladder because both forces pass through the axis.  Also for equilibrium,

Substituting Eq. 2 into Eq. 1:

23.

(a) Forces are shown in Fig. 10 above. The forces in addition to those shown in the statement of the problem are the attraction of the earth for the board = mg = 40 N and the force of the fulcrum on the board. For equilibrium, the vector sum of the forces = 0.   290 N - 50 N - 100 N - 40 N - 100 N = 0.

(b) Taking the axis at O, the net torque =

            (1 m) 50 N + (0.5 m) 100 N - (1 m)100 N = 0

Neither the weight of the board nor the force of the fulcrum on the board produce rotation about O because they pass through the axis.

(c) About O", the sum of the torques =

           -0.5 m(100 N) –1 m(40 N) – 2 m (100N) + 1 m(290 N)
           - 50 N-m - 40 N-m - 200 N-m + 290 N-m = 0.

The force of 50 N at O" produces no torque. It does not make any difference where you take the axis as long as you stick with it through a single calculation.

24.

(a)

(b) Take torques about the point where the object touches the floor. Then neither the frictional force nor the F_normal contribute to the torque because these forces pass through the axis. The perpendicular component of the distance r_W from the axis to the point of application of F_W is shown in Fig. 5 above to be 1.50m.  The perpendicular component of the distance r_mg from the axis to the point of application of mg is shown in Fig. 5 to be 0.50m.
Στ = 0 = +(0.5 m)30 N - (1.50 m)(F_W).
F_W = 10 N.

(c) σF_x = 0 = FW - f or FW = 10 N = f

(d) µ = f/F_normal = 10 N/30 N - 1/3