Hollow shere energy and velocity

  • Thread starter tebes
  • Start date
  • #1
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Homework Statement


A hollow sphere of radius 0.150 m, with rotational inertia I = 0.0513 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 25.3° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 54.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.10 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?


Homework Equations


Beforehand, I use rotational inertia to get mass already.



The Attempt at a Solution


I got a and b.
for (c),
I let
mgh = 54 J
solve for h.
h = 1.61 m
then, sin θ = opp/hyp
solve for hyp , and i got 3.77 m
(3.77 + 1.1 ) sin 25.3° = 2.08 m
mg( 2.08 - 1.61) = 15.75 J
but the answer is wrong. Can someone point out my mistake. Thank you.

For (d)
I used conservation of energy.
let mgh = 1/2 Iw^2 + 1/2 mv^2
then, I transformed the equation above into
mgh = 5/6 mv^2
solve for v.
I got 4.34 m/s
But the answer is wrong. Can someone point out my mistake. Thank you.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
When the sphere has moved 1.10 m up the incline from its initial position, what are (c) its total kinetic energy
So why do you still use 54 J when it has lost mg*1.1*sin 25.3° of that energy?
Begin with the 1.1 m, find the corresponding height and energy. Subtract from 54.
 

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