# Ball rolling down incline

• df102015

## Homework Statement The image above shows a basketball (a thin spherical shell I=⅔ mR^2) rolling don an incline of height 8.4 m. If the ball is already rolling with an initial linear speed of 3.0 m/s then what will be the final linear speed when it rolls off the incline?

## Homework Equations

I = 2/3 m R^2
KE = 1/2 m v^2 + 1/2 I ω^2
ω = v / R
PE = m g h

## The Attempt at a Solution

I know that i am missing some equations but honestly i don't know where to start other than substituting for ω and I.

KE = 1/2 m v^2 + 1/2 (2/3 m R^2) [(v / R)^2]

i do not think that i need the potential energy equation, but i mentioned it just incase. Also, how do i cancel out the mass and radius since they are not given in the problem?

You have all the equations. You just need the concept. Is anything conserved in this problem?

• df102015
You have all the equations. You just need the concept. Is anything conserved in this problem?
Energy, so would i set KE=PE ?

Energy, yes. But there is no law that says that KE should always equal PE. What about total mechanical energy: E = KE + PE? What can you say about E as the ball rolls down the slope?

• df102015
Hint:
Total Initial Energy = Total final energy
(As the ball rolls without slipping no heat is lost due to friction)

• df102015
Okay so would i use the equation that i have already set up? ...
KE = 1/2 m v^2 + 1/2 (2/3 m R^2) [(v / R)^2]
And would i add PE (mgh) and set equal to ME?

Hint:
Total Initial Energy = Total final energy
(As the ball rolls without slipping no heat is lost due to friction)
But how do i set the equations up so that the variables i need cancel and everything else remains?

But how do i set the equations up so that the variables i need cancel and everything else remains?
Do what you proposed in post #6 (add in PE), write the KE+PE expressions for each of the starting and ending circumstances, and set them equal.

• df102015
Do what you proposed in post #6 (add in PE), write the KE+PE expressions for each of the starting and ending circumstances, and set them equal.
Okay i think i have it...
1/2 m Vinitial^2 + 1/2 (2/3 m R^2) [(Vinitial / R)^2] + m g h = 1/2 m Vfinal^2 + 1/2 (2/3 m R^2) [(Vfinal / R)^2]

Is this correct?

Okay i think i have it...
1/2 m Vinitial^2 + 1/2 (2/3 m R^2) [(Vinitial / R)^2] + m g h = 1/2 m Vfinal^2 + 1/2 (2/3 m R^2) [(Vfinal / R)^2]

Is this correct?
Yes.

• df102015
Yes.
Thanks! I got the right answer!