Question

A ball of mass 'm" is rotated in a vertical circle with constant speed. The difference in tension at thetop and bottom would be1) 6 mg2) 5 mg3) 2 mg4) mg​

Answer 1

Answer:

$\boxed{\mathfrak{(1) \ 6 mg}}$

Explanation:

Mass of ball is given as 'm'

Let velocity at bottom/lowest point of vertical circle be 'v'

Tension in the string at bottom/lowest point of vertical circle is:

$\rm T_L = \dfrac{m {v}^{2} }{r} + mg$

Tension in the string at top/highest point of vertical circle is:

$\rm T_H = \dfrac{m {v}^{2} }{r} - 5 mg$

So, difference in tension at the top & bottom will be:

$\rm \implies T_L - T_H = \dfrac{m {v}^{2} }{r} + mg - ( \dfrac{m {v}^{2} }{r} - 5 mg) \\ \\ \rm \implies T_L - T_H = \cancel{\dfrac{m {v}^{2} }{r} } + mg \cancel{ - \dfrac{m {v}^{2} }{r} } + 5 mg \\ \\ \rm \implies T_L - T_H = 6mg$