1. Abowler throws a bowling ball of radius R 11.0 cm down the lane with initial

Question

1. Abowler throws a bowling ball of radius R 11.0 cm down the lane with initial speed vo 8.50 m/s, the ball is thrown in such a way that it skids for a certain distance before itstarts to roll. It not rotating at all when it first hits the lane, its motion being pure trans slation. The coefficient of kinetic friction between the ball and the lane is o.21. (a) For what length oftime does the ball (b) How far down the lane does it skid? (c) How many revolutions does it make before it skid? starts to roll? (d) How fast is it moving when it starts to roll? 2. A 5.13 kg object moves on a horizontal frictionless surface under the influence of a spring with force constant 9.88 N/cm. The object is displaced 53.5 cm and given an initial velocity of 11.2 m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) the initial potential energy of the system, (c the initial kinetic energy, (d) the am plitude of the motion. 3. A solid cylinder attached to a horizontal massless spring can roll without slipping along a horizontal surface as in the figure. The force constant kof the spring is 2.94 NVcm If the system is released from rest at a position in which the spring is stretched by 23.9 cm, find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the center of mass of the cylinder executes simple harmonic motion with a period: T 2 nV3M/2k, where Mis the mass of the cylinder. 4. A52.3 kg uniform square sign, 1.93 m on a side, is hung from a 2.88 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.12 m above the point where the rod is fixed to the wall, as shown in the figure. (a) Find the tension in the cable. (b) Calculate the horizontal and vertical components of the force exerted by the wall on the rod 5. A particle of mass m, moving with speed Vi, collides head on with a particle of mass m2, initially at rest, in a completely inelastic collision. (a) What is the kinetic energy of the system before the collision? (b) What is the kinetic energy of the system after the collision? (c) What fraction of the original kinetic energy is lost? (d) Let vom be the velocity of the center of mass of the system. View the collision from a primed reference frame moving with the Repeat parts (a), (b), and (c) as seen by an center of mass so that v u vi-Vom and Van observer in this reference frame. Is the kinetic energy lost the same in each case? Explain 6. A cue ball strikes another ball initially at rest. After the collision, the cue ball moves at 3.50 m/s along a line making an angle of65.00 with its original direction of motion. The second ball acquires a speed of 6.75 m/s. Using conservation of momentum, find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball, and (b) the original speed of the cue ball A man stands on a platform that is rotating with an angular speed of 1.22 revls; his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is 6.13 kgm If by moving the weights the man decreases the rotational inertia to 1.97 kgm2, a) what is the resulting angular

Explanation / Answer

7. (a) The resulting angular speed of a platform will be given as :

we know that,   L1 = L2

I1 w1 = I2 w2

w2 = [(6.13 kg.m2) (1.22 rev/s)] / (1.97 kg.m2)

w2 = 3.79 rev/s

(b) The ratio of new kinetic energy to the original kinetic energy which is given as :

(K.E)new / (K.E)original = (1/2) I1 w12 / (1/2) I2 w22

(K.E)new / (K.E)original = I1 w12 / I2 w22

(K.E)new / (K.E)original = [(6.13 kg.m2) (1.22 rev/s)2] / [(1.97 kg.m2) (3.79 rev/s)2]

(K.E)new / (K.E)original = [(9.123892) / (28.297277)]

(K.E)new / (K.E)original = 0.322

12. A sphere, a cylinder and a hoop, each of radius R and mass M, start from rest and roll down the same incline.

(a) Which object gets to the bottom first?

Cylinder

(moment of inertia greater than sphere & hoop)

(b) Yes, our answer depends on the mass or radius.

Mathematically, we have

Sphere :   I = (2/5) M R2

Cylinder :   I = (1/2) M R2

Hoop :    I = M R2

15. Given that,

mass of command module = mc

mass of engine, me = 4 mc

velocity of the command module, vc = ?

using conservation of momentum, we have

M vi = mc vc + me [vc - (125 km/hr)]

(mc + me) vi = mc vc + me [vc - (125 km/hr)]

(mc + 4 mc) (3860 km/hr) = mc vc + (4 mc) [vc - (125 km/hr)]

(5 mc) (3860 km/hr) = mc vc + (4 mc) vc - (500 km/hr) mc

(5 mc vc) = [(19300 km/hr) + (500 km/hr)] mc

vc = (19800 km/hr) / 5

vc = 3960 km/hr

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