A solid 0.7150 kg0.7150 kg ball rolls without slipping down a track toward a vertical loop of radius?=0.7350 mR=0.7350 m. What minimum translational speed? minvmin must the ball have when it is a height?=1.151 mH=1.151 m above the bottom of the loop in order to complete the loop without falling off the track?

Question

Lance Larson

Physics

13 September, 15:59

A solid 0.7150 kg0.7150 kg ball rolls without slipping down a track toward a vertical loop of radius?=0.7350 mR=0.7350 m. What minimum translational speed? minvmin must the ball have when it is a height?=1.151 mH=1.151 m above the bottom of the loop in order to complete the loop without falling off the track?

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Answers (1)

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Edith Sanford · Answer 1

3.66 m / s

Explanation:

Initially the ball is at a height of 1.151m, so its potential energy

= mx9.8x1.151 = 11.2798m J

If V be its required velocity there then its kinetic energy = 1/2 m V²

Total mechanical energy = 11.2798m + 1/2 m V²

The potential energy at the top of loop =

= m x 9.8 x 2 x. 7350 = 14.406m

if v be the velocity there, the centripetal force must balance the weight there so mv²/r = mg; v² = gr

kinetic energy = 1/2 m v² = 1/2 m gr =.5 x m x 9.8 x. 7350 = 3.6m

Applying conservation of energy,

11.2798m + 1/2 m V² = 14.406m + 3.6 m

11.2798 +.5 V² = 14.406 + 3.6

V² = 13.4524

V = 3.66 m/s