Question

A horizontal turntable (disc) of mass = 5kg, radius = 1.80m, is spinning about its center at the rate of 5 rad/s. A lump of mass 0.800 kg falls on it vertically and sticks to it at a distance of 1.20 m from the center.

A) Calculate the new angular speed of the disc.

B) What is the new rotational kinetic energy of the disc+lump

Answer 1

A) 4.38 rad/s

B) 88.75 J

Step-by-step explanation:

A)

M = mass of the disc = 5 kg

R = radius of the disc = 1.80 m

I = initial moment of inertia of the system = (0.5) MR² = (0.5) (5) (1.80)² = 8.1 kgm²

m = mass of the lump = 0.800 kg

r = distance of the lump from axis of rotation = 1.20 m

I' = initial moment of inertia of the system = I + m r² = 8.1 + (0.800) (1.2)² = 9.252 kgm²

w' = new angular speed

w = initial angular speed = 5 rad/s

Using conservation of angular momentum

I w = I' w'

(8.1) (5) = (9.252) w'

w' = 4.38 rad/s

B)

new rotational kinetic energy of the disc+lump is given as

KE' = (0.5) I' w'²

KE' = (0.5) (9.252) (4.38)²

KE' = 88.75 J