What is its linear speed at the bottom of the incline?

  • #1

Homework Statement


A spherical object with a moment of inertia of 0.584mr2 starts from rest rolling down a 2.25-m high incline. If the sphere is rolling without slipping


Homework Equations


I the best equation to use for this problem is k(initial)+u(initial)=k(final)+u(final)


The Attempt at a Solution


I stretch the equation 1/2mv^2(initial)+1/2Iω^(initial)+mgh(initial)=1/2mv^2(final)+1/2Iω^2(final)+mgh(final).

since the object started from rest the initial kinetic energy and the final potential energy is zero, which leads me to this equation

mgh(initial)=1/2mv^2(final)+1/20.584mr^2ω^2(final). As I continue reduce the equations I round up with this.

gh(initial)=0.792v^2(final)

v=sqrt(9.81m/s^2)(2.25m)/0.792

the linear speed I came up with was 5.279m/s.

Did I do this right?
 

Answers and Replies

  • #2
haruspex
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v=sqrt(9.81m/s^2)(2.25m)/0.792
The parentheses aren't written correctly there, but the method's fine and I agree with your final answer.
 
  • #3
The parentheses aren't written correctly there, but the method's fine and I agree with your final answer.
How would I write the parentheses the correct way?
 
  • #4
haruspex
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How would I write the parentheses the correct way?
sqrt((9.81m/s^2)(2.25m)/0.792)
 

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