
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
a = (v  v_{o})/(t  0)
or
v = v_{o }+ at.

(b)

Constant angular acceleration α
equals the change in angular velocity divided by the
time to change the velocity
α = (ω  ω_{o})/(t  0)
or
v = v_{o }+
at.



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}
(b)

Θ(t)
= Θ_{o}
+ ω_{o}t
+ 1/2 αt^{2}


Θ(4.0s)
= 0 + ( π s^{1})(4.0s)
+ 1/2 (4 πs^{2})(16s^{2})


= 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.

(a) ω(t)
= dΘ/dt = d(b + ct + et^{2})/dt
= c + 2et. ω(t_{1})
= c + 2et_{1
} (b) α(t) = dω/dt
= d(c + 2et)/dt= 2e = constant = α(t_{1})
(c) v(t_{1}) = ω(t_{1})r
= (c + 2et_{1})r
(d) tangential acceleration, a(t_{1}) = α(t_{1})r
= 2er
(e) centripetal or radial acceleration(t_{1}) ={ v(t_{1})}^{2}/r
= (c + 2et_{1})^{2}r^{2}/r
= (c + 2et_{1})^{2}r


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,
v_{1} = s_{1}/t, and for point 2, v_{2}
= s_{2}/t. Since s_{2 }> s_{1},
v_{2 }> v_{1}. This is also shown
from v_{1} = ωr_{1
}and v_{2} = ωr_{2}.
Again since r_{2 }> r_{1}, v_{2
}> v_{1}.


7.

Angular acceleration α
= (ω_{10s}  ω_{o})/(3 s  0) =(6.0 rad/s  0)/(3 s) = 2
rad/s^{2}. Angle turned through from t = 0 to t =
10 s = ω_{o}t + 1/2 αt^{2}
= 0 +
1/2(2 rad/s^{2})(100 s^{2}) = 100 rad. ω(10
s) = ω_{o }+ αt
= 0 + (2.0 rad/s^{2})(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 αt^{2 }= 100
rad + 20 rad/s(10 s) + 0 = 300 rad.



9.

(a) I = mL^{2} + m(4L^{2}) = 5mL^{2}.
(b) In general, K = 1/2 Iω^{2}.
For object closest to the axis O,
K
= 1/2 mL^{2}ω^{2}.
For object farthest from O,
K
= 1/2 4mL^{2}ω^{2}.
Total
K = 1/2 L^{2}ω^{2}
(1 + 4) = 5/2(L^{2}ω^{2}).


10.

τ
= r x F
τ_{1
}= r_{1}F_{1 }sin 30^{0} = 2.0
m(1.5 N)(0.5) = 1.5 Nm
Point the fingers of your right hand in the direction of r_{1}
and rotate it into F_{1} and your thumb points
into the page. τ_{1
}is into the page.
τ_{2
}= r_{2}F_{2 }sin 30^{0} = 1.0
m(1.0 N)(0.8) = 0.5 Nm
Point the fingers of your right hand in the direction of r_{2}
and rotate it into F_{2} and your thumb points
out of the page. τ_{2
}is out of the page. Resultant torque τ= (1.5  0.8)Nm = 0.7 Nm into the page.


11.

(a) Moment of inertia I of a rod about
one end is 1/3 mL^{2}.
K = 1/2 Iω^{2}
= 1/2(1/3mL^{2})ω^{2}.
(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 (mL^{2})ω^{2}.
Change in height of center of mass = (L^{2}ω^{2}/6g).


12.

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

+

K_{i}

=

U_{f}

+

K_{f}

mgL/2

+

0

=

0

+

1/2 Iω^{2}








ω
= (mgL/I)^{1/2} = (mgL/1/3 mL^{2})^{1/2
}= (3g/L)^{1/2}

v_{top
of pencil } = ωL
= (3g/L)^{1/2}L = (3gL)^{1/2}



13.

(a)

See Fig. for #13 to the right,



τ_{about
pivot }= I_{about pivot }α


L/2(Mg)sin 90^{0
}= 1/3 ML^{2} α


α
= 3/2 (g/L)

(b) a = α
L = 3/2(g/L) (L) = 3g/2



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 =
{R^{2 } (R  h)^{2}}^{1/2} ={R^{2
} R^{2} + 2Rh  h^{2}}^{1/2}
= {2Rh  h^{2}}^{1/2}. The torque due
to mg = {2Rh  h^{2}}^{1/2}mg 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  h^{2}}^{1/2}mg,
or F must be slightly greater than
{2Rh  h^{2}}^{1/2}mg/(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 90^{0} = rmv.
(b) Since v = ωr, L =
(mr^{2})ω = Iω.


17.




(a)

L = rp sin r,
p = p(r sin
r, p) = pr_{}



(b)

L = r(p sin r,
p) = r p_{}



18.





L = r x mv


L is into the page


L = rmv sin r


v = mv(r sin r,
v) = mvb





19.





L = r x mv


L is out of the page for
both particles
(See Fig. 7 to the right)


L = rmv sin r


v = rmv sin 90^{0}


Total L = 0.50 m(5.0 m/s)(3.0 +
4.0) kg


=
17.5 kgm^{2}/s





20.





τ
= r x F


τ
= r
F sin r, F



For the satellite orbiting about the
earth as its axis, r,
F = 180^{0}.
sin 180^{0 }= 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 P_{1} = 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,
U_{i } + K_{i }
= U_{f} + K_{f}

Mgh + 0 = 0 + 1/2 Mv^{2}
+ 1/2 Iω^{2 }=
0 + 1/2 Mv^{2} + 1/2(2/5MR^{2})(v/R)^{2}

Mgh = 7/10 Mv^{2} or v
= (10gh/7)^{1/2}

(b)
(F_{net})_{x }=
ma


Mg sin Θ
 f = Ma

(Equation
1) 
Take the torques about the center of
mass. Since the weight passes through this axis, it produces
no torque.
τ
= Iα


Rf sin 90^{0} = (2/5 MR^{2})(a/R);
f = 2/5 Ma

(Equation
2) 
(i)


Substituting Eq. 2 into Eq. 1:
Mg sin Θ  2/5 Ma
= Ma or a = 5g sin Θ/7

(ii)


From Eq. 2, f = 2/5 Ma = (2/5)M(5g
sin Θ/7) = 2 Mg
sin Θ/7

(iii)


v^{2} = v_{o}^{2}
+ 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.

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,
σ
F_{x} = 0

and

σ
F_{y} = 0

f  F_{wall }= 0
f = F_{wall}


N  W = 0




N
= W but f = µN

so f
= µW and µW = F_{wall} 51.3^{0} (Equation
2)

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

Θ
= 51.3^{0}



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 Nm  40 Nm  200 Nm + 290 Nm = 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)

σ
F_{y} = 0


F_{normal}  mg = 0

or 
F_{normal} = mg = 30 N

(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

