a force applied tangentially to the disk...?

Question

A horizontal 795 N merry-go-round is a solid disk of radius 1.40m, started from rest by a constant horizontal force of 50.3 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 2.96s.

soooo stuck i think you need to find torque but i really have no idea

the first 2 answers are wrong... they are apparently not even close... any more ideas?

Answer 1

Correct, first find the torque:

t = F*r = 50.3 N * 1.4 m = 70.42 N*m

Then use Newton's 2nd law for rotational motion:

Sum of torques = I * alpha where I is the moment of inertia and alpha is the angular accelaration.

Mass = 795 N / 9.81 m/s^2 = 81.04 kg

I for a solid disk = .5 * mass * radius^2
= .5 * 81.04 kg * (1.4 m)^2 = 79.42 kg*m^2

alpha = torque / I = 70.42 N*m / 79.42 kg*m^2 = 0.887 radians/s^2

Then use a constant acceleration equation to find the angular velocity, w:

w = w(initial) + alpha * t = 0 + .887 rad/s^2 * 2.96 s. = 2.63 rad/s.

The kinetic energy = .5*I*w^2 = .5 * 79.42 kg*m^2 * (2.63 rad/s)^2 = 275 Joules

Answer 2

The torque of the force is
50.3*1.40

The moment of inertia of the merry go round is
(795/9.81)*1.4^2/2

torque is related to angular acceleration as
Torque = I * alpha
solve for alpha

50.3*1.40/((795/9.81)*1.4^2/2)
and plug into
w=alpha * t
t=2.96 seconds
once you have w, compute the KE as
KE=.5*I*w^2

α=0.887 rad/s^2
w=0.887*2.96
w=2.66 rad/sec

KE=.5*((795/9.81)*1.4^2/2)*2.66^2
KE=281 J

j


j