How is Rotational Kinetic Energy Calculated for a Merry-Go-Round?

AI Thread Summary
To calculate the rotational kinetic energy of a merry-go-round, the problem involves a horizontal merry-go-round with a weight of 813 N and a radius of 1.28 m, started by a constant tangential force of 66 N. The equations for weight, force, tangential velocity, angular speed, and moment of inertia for a solid cylinder are utilized in the calculations. The user initially found the tangential acceleration and subsequently calculated the tangential velocity and angular speed, leading to an attempt to compute the rotational kinetic energy. However, the final answer of 94.31353 Joules was marked incorrect, indicating a potential oversight in considering torque. The discussion emphasizes the importance of torque in the calculations for accurate results.
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Homework Statement



A horizontal 813 N merry-go-round of radius 1.28 m is started from rest by a constant horizontal force of 66 N applied tangentially to the merry-go-round. The acceleration of gravity is 9.8 m/s^2. Assume it is a solid cylinder. Find the kinetic energy of the merry-go-round after 2.68 s.

Homework Equations


Weight = mg
F=ma
Vf=Vi + at
tangential velocity=radius*angular speed
Kr = 1/2 (moment of inertia*angular speed^2)
moment of inertia = 1/2 Mass*Radius^2 for solid cylinder

The Attempt at a Solution


Well, using the constant force applied and the weight of the merry-go-round, I found the tangential acceleration: a = F * g / weight...Then I solved for the tangential velocity with Vf = a * t since it started from rest...then I solved for the angular speed:
w = tangential velocity / radius. Then plugged all the numbers I found to solve for moment of inertia and rotational kinetic energy. My final answer is 94.31353 Joules. I entered it in the computer and I got it wrong. I'm not sure where my error is. pls advice. Thanx!
 
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oh. I should have considered torque...
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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