(10%) Problem 4: A bowling ball m 2.8 kg drops from a height h semi-circular tub

Question

(10%) Problem 4: A bowling ball m 2.8 kg drops from a height h semi-circular tube r 5.8 m rests centered on a scale. 82 m. A 0 Scale Otheexpertta.com 50% Part (a) write an expression for the reading of the scale when the bowling ball is at its lowest point, in terms of the variables in the problem statement. Grade Summary Potential 100% Submissions Attempts remaining: S 0 per attemp detailed vlew 0 I give up deduction per fedback Submit Hint Hints: 0% deduction per hint. Hints remaining:- Feedback: 50% Part (b) what does the scale read. in newtons? CI O 201 Epen TA LC

Explanation / Answer

Using conservation of energy, mgh = 1/2 mv2                    ...(1)
Where mgh is the potential energy at the top, where kinetic energy is zero. Potential energy is zero at the bottom.
From (1),
mv2 = 2mgh
Dividing with r,
mv2/r = (2mgh) / r             ..(2)

At the bottom of the loop, the ball have centripetal force in addition to its weight
Force on the scale, W = mg + mv2/r
Substituting (2),
W = mg + (2mgh)/r

b)
Substituting values in the above equation,
W = 2.8 x 9.8 + [2 x 2.8 x 9.8 x 8.2] / 5.8
= 105.03 N

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