Rotational Kinetic Energy of sphere

Click For Summary
SUMMARY

The discussion centers on the rotational kinetic energy of a solid sphere rolling up an incline. The sphere, with a mass of 4 kg and a radius of 0.5 m, has a moment of inertia defined as I = 2/5(m)R^2. The total kinetic energy at the bottom of the incline is calculated to be 70 J, and the sphere travels 2.86 m up the incline before coming to rest. It is confirmed that the distance traveled does not depend on the mass of the sphere, as mass cancels out in the energy equations.

PREREQUISITES
  • Understanding of rotational and translational kinetic energy
  • Familiarity with the moment of inertia formula for solid spheres
  • Basic principles of energy conservation in physics
  • Knowledge of trigonometric functions related to incline problems
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about energy conservation principles in rolling motion
  • Explore the effects of friction in rolling without slipping scenarios
  • Investigate the relationship between mass, radius, and acceleration in rolling objects
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of rolling motion and energy conservation principles.

latitude
Messages
54
Reaction score
0

Homework Statement



A solid sphere of mass 4 kg rolls w/o slipping UP an incline with an angle of 30 degrees. the radius of the sphere is 0.5 m and its moment of Inertia is I = 2/5(m)R^2. At the bottom of the incline the center of mass of the sphere has a translational speed of 5 m/s.
a) What is the total kinetic energy of the sphere at the bottom of the incline?
b) How far does the sphere travel up the incline before coming to rest and starting to roll back down?
c) Does the answer to b) depend on the mass?



The Attempt at a Solution



a) K = Krotational + Ktranslational
= 1/2Iw^2 + 1/2mv^2
w = v/R = 5/0.5
= 1/2(2/5)(4)(o.5^2)(10^2) + 1/2(4)(5^2)
= 70 J

b) Kf + Uf = Ki + Ui
0? (Not sure about this, because not sure if there is still some rotational kinetic energy?) + mgh = 70 J + 0J
(5)(9.8)(h) = 70
h = 1.43 m

h = xsin30
1.43 = xsin30
x = 2.86 m

c) So this is where I screw up... because I know from my theory classes that the answer SHOULDNT depend on the mass because all spheres roll down the same regardless of mass... but mine does so I think it's wrong. Thanks everyone!
 
Physics news on Phys.org
b.) When the sphere starts to roll back down, it means that it has lost all its kinetic energy to potential energy

c.) no, because when you do b, you will notice that m cancels out from the equation. Do part b and you will udnerstand.
 
latitude said:
(Not sure about this, because not sure if there is still some rotational kinetic energy?)
Since the sphere is rolling without slipping, if it moves, it has to spin. That's why both the translational movement and rotation stops at the same time.

c) So this is where I screw up... because I know from my theory classes that the answer SHOULDNT depend on the mass because all spheres roll down the same regardless of mass... but mine does so I think it's wrong. Thanks everyone!

How do you know that your answer does depend on the mass? This is what happens if you plug in numerical values early in the problems. As mentioned by Oerg, do it taking m as the mass and see whether it cancels out.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
Replies
1
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
6
Views
1K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 5 ·
Replies
5
Views
933