A 0.30-kg rock is swung in a circular path and in a verticalplane on a 0.25-m-le

Question

A 0.30-kg rock is swung in a circular path and in a verticalplane on a 0.25-m-length string. At the top of the path, theangular speed is 12.0 rad/s. What is the tension in the string atthat point? 1) Draw a free body diagram and write the force equation for the rock at the top of the path. I figured the tension on top to be 7.9 N. The forces acting upon is on the top; tension and Gravity on the bottom pulling it down. Would that be the answer? 2) Draw a free body diagram and write the force equation for the rock at the BOTTOM of the circle. If the angular speed remains constant, whats is the tension in the strong at the bottom of the circle.

Explanation / Answer

Weight of the rock is 0.3 x 9.8 = 2.94 N. m v v / r = m (r w) (r w) / r, since v = r w. Or use m r w^2. 0.3 *(0.25 x 12)* (0.25 x 12) / 0.25 =10.8 N. Tension = (10.8 - 2.94) = 7.86 N

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