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**1.**A bowling ball with a mass of 6.95 kg and a radius of 0.194 m starts from rest at a height of 2.21 m and rolls down a 49.2 degree slope. What is the translational speed of the ball when it leaves the incline?

**2.**Unsure! Here's what I've got, though:

ω = v/r

mgh=1/2mv

^{2}+1/2Iω

^{2}

And so: ω= √(2mgh/(mr

^{2}+I))

I of a solid sphere = 2/3(m)(r

^{2})

## The Attempt at a Solution

I = 2/3(6.95)(.194

^{2}) = 0.174

ω= √(2(6.95)(9.8)(2.21)/((6.95)(.194

^{2}+0.174)) = √(301.05)/(0.436) = 26.29

v = ωr

v = 26.20 X 0.194 =

*5.1*

Alas, 5.1 is

*not*the correct answer. I'm sure I need to factor in the angle somewhere, but I don't know how! Help, please!