A stone of mass 200 g attached to a string is made to revolve in a horizontal circle of radius 1.5 m at a steady speed of 5 m/s. Calculate the tension in the string at the bottom of the circle.
5.3 N
3.3 N
2.0 N
1.3 N
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Discussions (5)
The answer provided is wrong. The description given in your solution is that of uniform vertical circle. In this case, the weight affects the tension at the bottom and at the top.
For horizontal circle, the tension is simoly equal to the centripetal force throughout the circle.
So,
T = Fc = mv^2/r = 3.3 N
Greetings, The Myschool Team.
The correct answer is B(3.3N) and not A(5.3N). According to New School Physics (pg. 25, 2016 Edition), the tension in the string provides the centripetal force i.e tension = centripetal force. Thus, the tension in the string is 3.3N which is the centripetal force. @Chris also made a valuable contribution to the subject matter. Thank you.
