B LEVEL PROBLEMS
1.Calculate the net torque about the axle of the wheel shown below. Assume that a frictional torque of 0.40 mN opposes the motion (F1= 35 N, F2 = 30 N, F3 = 20 N, r = 10 cm, and R = 20 cm). [1.1 m·N (clockwise)]

SOLUTION:

We assume clockwise motion, so the frictional torque is counterclockwise.If we take the counterclockwise direction as positive, we have 

            tnet= -rF1 + RF2 - RF3 + tfr

            = -(0.10 m)(35 N) + (0.20 m)(30 N) - (0.20 m)(20 N) + 0.40 m · N 

            =1.1 m · N (clockwise).

2.Calculate the moment of inertia of the array of masses shown below about (a) the vertical axis, and (b) the horizontal axis. Assume the connecting wires have negligible mass. About which axis will it be harder to accelerate this object? (m = 1.8 kg, M = 3.1 kg, h = 0.50 m, d1 = 0.50 m, and d2 = 1.50 m)[6.1 kg·m2, 0.61 kg·m2]

SOLUTION:

(a)For the moment of inertia about the y-axis, we have

            Ia=SmiRi2= md12 + Md12 + m(d2 – d1)2 + M(d2 – d1)2

            = (1.8 kg)(0.50 m)2+ (3.1 kg)(0.50 m)2 + 

            (1.8 kg)(1.00 m)2 + (3.1 kg)(1.00 m)2

            =6.1 kg · m2.

(b)For the moment of inertia about the x-axis, all the masses are the same distance from the axis, so we have

            Ib= SmiRi2= (2m + 2M)(0.5h)2

            = [2(1.8 kg) + 2(3.1 kg)](0.25 m)2= 0.61 kg · m2. 

3.A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 5.80 kg. Calculate (a) its moment of inertia about its center, (b) the applied torque needed to accelerate it from rest to 1500 rpm in 5.00 s if it is known to slow down from 1500 rpm to rest in 55.0 s.[ 1.99´10–3 kg·m2,6.82 ´10–2 m·N.]
SOLUTION:
(a)The moment of inertia of the solid cylinder is 

            I = ½ MR2 = ½(0.550 kg)(0.0850 m)2=1.99 ´ 10–3 kg · m2. 

(b)We can find the frictional torque acting on the wheel from the slowing-down motion:

            tfr= Ia1 = I(w1 –w01)/t1

                = (1.99 ´ 10–3 kg · m2)[0 – (1500 rpm)(2p rad/rev)/(60 s/min)]/(55.0 s) = – 5.67 ´10–3 m · N.

For the accelerating motion, we have

            tapplied + tfr = Ia2 = I(w2 –w02)/t2;

            tapplied – 5.67 ´ 10–3 m · N = (1.99 ´ 10–3 kg · m2)[(1500 rpm)(2p rad/rev)/(60 s/min) – 0]/(5.00 s),

which gives

            tapplied = 6.82 ´ 10–2 m · N.

4.The radius of a roll of toilet paper is 7.6 cm and its moment of inertia is 2.9 x 10-3 kgm2.A force of 3.2 N is exerted on the end of the roll for 1.3 s but the paper does not tear so it begins to unroll. A constant frictional torque of 0.11 Nm is exerted on the roll which gradually brings it to a stop. Assuming the thickness of the paper is negligible, calculate (a) the length of the paper during the 1.3 s in which the force was applied, and (b) the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving. [2.9 m,3.6 m] 
SOLUTION:
We choose the clockwise direction as negative
(a)With the force acting, we write St= Ia about the axis from the force diagram for the roll:
            Fr – tfr = Ia1;
            (3.2 N)(0.076 m) – 0.11 m · N = (2.9 ´ 10–3 kg · m2)a1 , 

which gives a1  = 45.9 rad/s2.

We find the angle turned while the force is acting from 

            q1 =w0t + ½ a1 t1 2

            = 0 + ½ (45.9 rad/s2)(1.3 s)2 = 38.8 rad.

The length of paper that unrolls during this time is

            s1 = rq1 = (0.076 m)(38.8 rad) =2.9 m.

(b)With no force acting, we write St= Ia about the axis from the force diagram for the roll:

            –tfr = Ia2;

            – 0.11 m · N = (2.9 ´ 10–3 kg · m2)a2, 

which gives a2 = – 37.9 rad/s2.

The initial velocity for this motion is the final velocity from part (a):

            w1 =w0 + a1t1 = 0 + (45.9 rad/s2)(1.3 s) = 59.7 rad/s.

We find the angle turned after the force is removed from 

            w22=w12+ 2a2q2;

            0 = (59.7 rad/s)2 + 2(– 37.9 rad/s2)q2, which gives q2 = 47.0 rad.

The length of paper that unrolls during this time is

            s2 = rq2 = (0.076 m)(47.0 rad) =3.6 m.

5.A merry-go-round has a mass of 1640 kg and a radius of 8.20 m. How much net work is required to accelerate it from rest to a rate of 1 revolution in 8.0 s? Assume that it is a solid cylinder. [1.70 ´104 J] 
SOLUTION:
The work done increases the kinetic energy of the merry-go-round:

            W= Dke = ½ Iw2 – 0 = ½ (½ Mr2)(Dq/Dt)2

            =((1640 kg )(8.20 m)2[(1 rev)(2p rad/rev)/(8.00 s)]2 =1.70 ´ 104 J.

6.Say you have an Atwood machine with masses m1 = 18.0 kg, and m2 = 26.5 kg. The pulley is a solid cylinder of radius 0.260 m and mass 7.50 kg. Initially, m1 is at rest on the ground and m2 is 3.00 m above the ground. If the system is subsequently released, use energy conservation to calculate the speed of m2 just before it strikes the ground. Assume the pulley is frictionless. [3.22 m/s] 
SOLUTION:
For the system of the two blocks and pulley, no work will be
done by nonconservative forces.The rope ensures that each block has the same speed v and the angular speed of the pulley is w = v/R0.We choose the reference level for gravitational potential energy at the floor.
The rotational inertia of the pulley is I = ½ MR02.
For the work-energy principle we have

            Wnet = Dke + Dpe;

            0 = [( ½ m1v2 + ½m2v2 + ½Iw2) – 0] + m1g(h – 0) + m2g(0 – h);

            ½m1v2 + ½m2v2 + ½ (½MRo2)(v/Ro)2 = (m2 – m1)gh;

            ½ [m1 + m2 + ½M]v2 = (m2 – m1)gh;

            ½ [18.0 kg + 26.5 kg + ½ (7.50 kg)]v2 = 

            (26.5 kg – 18.0 kg)(9.80 m/s2)(3.00 m), which gives

            v =3.22 m/s.