A ball of mass M = 1.40 kg and radius r = 4.70 cm is attached to one end of a th

Question

A ball of mass

M = 1.40 kg

and radius

r = 4.70 cm

is attached to one end of a thin, cylindrical rod of length

L = 12.0 cm

and mass

m = 0.800 kg.

The ball and rod, initially at rest in a vertical position and free to rotate around the axis shown in the figure below, are nudged into motion.

(a) What is the rotational kinetic energy of the system when the ball and rod reach a horizontal position?
J

(b) What is the angular speed of the ball and rod when they reach a horizontal position?
rad/s

(c) What is the linear speed of the center of mass of the ball when the ball and rod reach a horizontal position?
m/s

(d) What is the ratio of the speed found in part (c) to the speed of a ball that falls freely through the same distance?

=

Vc/Vfreefall

Explanation / Answer

given

M = 1.4 kg

r = 4.7 cm = 0.047 m

L = 12 cm = 0.12 m

m = 0.8 kg

a. intiial poitential energy of the system wrt horizontal , PE = Mg(L + r) + mgL/2

PE = 2.764458 J

this PE will be converted to kinetic energy when the bar reaches the horizontal position

so, rotaitonal KE = 2.764458 J

b. let the angular speed at this moment be w

then

0.5*(M(L + r)^2 + mL^2/12)*w^2 = 2.764458

w = 11.756 rad/s

c. distnace of center of mass form the hinge = x

then

(m + M)x = M(r + L) + mL/2

x = 0.10709 m

hence

linear speed of com = x*w = 1.2589 m/s

d. speed gained when the ball falls freely through distance L + r = sqroot(2*g*(L + r))

v' = 1.6211 m/s

ratio = 0.7764

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