How to solve rotational work and energy problem?

AI Thread Summary
The discussion focuses on solving a rotational work and energy problem involving a basketball and a can of frozen juice rolling down a hill. The initial total mechanical energy is equated to the final total mechanical energy, leading to the equation mgh = 1/2mv^2 + 1/2IW^2. The relationship between tangential velocity and angular velocity is clarified, confirming that tangential velocity equals translational velocity when rolling without slipping. The height of the hill is calculated to be 3.18 meters. The participants confirm their understanding of the concepts involved in the problem.
Ailiniel
Messages
5
Reaction score
0

Homework Statement



Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 6.12 m/s. Ignore frictional losses.

Homework Equations



(a) What is the height of the hill?

(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

The Attempt at a Solution



Initial Total mechanical energy = Final Total mechanical energy
1/2mv^2 + 1/2IW^2 + mgh = 1/2mv^2 + 1/2IW^2 + mgh
Initial KE = 0 and final PE = 0
mgh = 1/2mv^2 + 1/2IW^2
gh = 1/2v^2 + 1/2IW^2

I am stuck with 2 unknown variables W and h.
 
Physics news on Phys.org
Hint: Assuming each object rolls without slipping, what's the relationship between v and W?

Also: What's the rotational inertia of each object?
 
mgh = 1/2mv^2 + 1/2IW^2
Tangent Velocity = rW
W=Tangent V / r

Substitute W and solve for h.
h = (1/2V^2 + 1/3(Tangent V)^2) / g

Tangent V = translational V

h=3.18

I got it now, Thanks!
 
Last edited:
Is Tangent velocity the same as translational velocity?
Yes, because it rolls without slipping.
 
Ailiniel said:
I got it now, Thanks!
Good!
 
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...
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...
Back
Top