Question

In: Physics

A hollow sphere of radius 0.170 m, with rotational inertia I = 0.0150 kg

Expert Solution

Part A)

I for a hollow sphere is (2/3)mr²

KEr = .5Iw² and w = v/r.

Thus KEr = .5(2/3)(mr²v²/r²) which simplifies to (1/3)mv²

KEt = (1/2)mv²

Total KE = (1/2 + 1/3)mv² = (5/6)mv²

The rotational portion is (1/3)/(5/6)

That is 2/5 of the KE (which is .40 or 40%)

Part B)

At the initial position

KE = (5/6)mv²

To find m, apply I = (2/3)mr²

(.015) = (2/3)(m)(.17)²

m = .779 kg

6.9 = (5/6)(.779)(v²)

v = 3.26 m/s

Part C)

Total KE = 6.9 - mgh

There are two ways to interpret this. Is the height .710 m or is the .710 the distance along the incline meaning that the height is .71(sin 38) = .437 m. I will give you both answers...

KE = 6.9 - (.779)(9.8)(.71) = 1.48 J or...

KE = 6.9 - (.779)(9.8)(.71)(sin 38) = 3.56 J

Part D)

Also two possible answers depending on what is meant by the .71 m. Is that straight up, or along the incline.

The two answers are...

1.48 = (5/6)(.779)v²

v = 1.51 m/s or...

3.56 = (5/6)(.779)v²

v = 2.34 m/s

genius_generous answered 18 hours ago

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