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Chapter 8: Q. 68 (page 203)

The father of Example 8.2 stands at the summit of a conical hill as he spins his 20 kg child around on a 5.0 kg cart with a 2.0-m-long rope. The sides of the hill are inclined at 20o颅. He
again keeps the rope parallel to the ground, and friction is negligible. What rope tension will allow the cart to spin with the same 14 rpm it had in the example?

Short Answer

Expert verified

The tension on the rope is 185.4 N.

Step by step solution

01

Given information

Total mass of cart and child, m=20+5=25 kg
Radius of the rope, r=2 m
Angle of the hill, 胃=20o
Angular speed, N=14 rpm

02

Explanation

First draw the Free body diagram from the given information

Angular velocity is calculated by 蝇=2蟺n60

Substitute the values, we get

蝇=2蟺n60=2蟺rad(14rpm)60sec=1.47rad/s

From the figure we can equate the horizontal forces to calculate tension

T=mgsin20o+mr蝇2cos20o

substitute the values, we get

localid="1649094922951" T=(25kg脳9.81m/s2)脳sin20o+(25kg脳2m脳(1.47rad/s)2cos20o T=185.40N

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