A ball [5 kg, 78 cm radius, I_com = 0.4mr^2] rolls without supping along a rigid

Question

A ball [5 kg, 78 cm radius, I_com = 0.4mr^2] rolls without supping along a rigid surface. At point A it has a center of mass speed of 420 cm/s. Point A is 20 m above point B. From point B it rolls along a level surface and through a vertical loop of uniform inner radius. Point C is one-quarter of the way through the loop and point D is at the top. a) Calculate the angular velocity of the ball when it reaches point B. b) At point D, the normal force acting on the ball from the track has a magnitude of three times the weight of the ball. What is the inner radius of the loop? You are restoring a grandfather clock. Your pendulum design consists of a long thin rod [750 g, length 1.2 m, I_com = (m/12)L^2 that passes through the middle of a solid disc [400 g, radius 6 cm, I_com = 0.5mr^2] One end of the rod is attached to a hinge and the center of the disc is 75 cm from the hinge. At t = 0 seconds, the pendulum is released from rest at an angle of 0.1 radians relative to the vertical. a) Calculate how many cycles the pendulum completes in one hour. b) Construct an expression for the tangential acceleration of the tip of the rod as a function of time. The blade on a particular table saw has is a thin disc with moment of inertia of 0.15 kgm^2. As the blade interacts with the wood, the blades' angular momentum decreases to zero as the blade gets jammed in the wood grain. L vector = [200 pi^2 - 20 pi t + 0.5 t^2] (- k) has units of Nms when time is measured in seconds. a) Through what angle does the blade turn before it jams? b) Calculate the net work done on the blade between t = 6pi and t = 12 pi seconds.

Explanation / Answer

consider question 1:

PE at point A = mgh = 5*9.81*20 = 981 J
KE at point A = 0.5mv^2 = 0.5*5*04.2^2 = 44.1 J
TE at point A = 1025.1 J

a) At point BE
PE of ball = 0 J
KE of ball = rotational KE + Translational KE = 0.5mv^2 + 0.5Iw^2
but for pure rolling
w = v/r
and, I = 0.4mr^2
so
KE = 0.5mv^2 + 0.5*0.4mr^2*v^2/r^2 = 0.7mv^2 = 0.7*5*v^2
from conservation of energy
3.5v^2 = 1025.1 J
v = 17.113 m/s
w = v/r= 17.113/0.78 = 21.94 rad/s
b) At point D, N = 3mg
let inner radius of the shell be R
then from conservation of energy
1025.1 = mgh*2*R + 3.5v^2

also, mv^2/r = N + mg
so, 5v^2/0.78 = 4*5*9.81
v = 5.532 m/s

R = 9.357 m

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