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

[jimmyboykun]

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?

 

[haruspex]

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.

 

[jimmyboykun]

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?

 

[haruspex]

How would I write the parentheses the correct way?

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

 

[Nickie]

Yes, your approach and calculations seem correct. You have used the appropriate equations and correctly applied the conservation of energy principle to solve for the linear speed at the bottom of the incline. Make sure to double-check your calculations and units to ensure accuracy. Good job!