A string 0.5 ~m long is used to whirl a 1 ~kg stone in a vertical circle at a uniform velocity of 5 ~m ~s^ -1 . What is the tension in the string when the stone is at the top of the circle?

Correct answer: C. 40.2 ~N The centripetal force needed to keep the stone moving at 5 ~m ~s^ -1 is F_c= (m v^2)/(r)= (1 ~kg)(5 ~ms^ -1 )^2 0.5 ~m = 50 ~N The weight of the stone is W=m g=(1 ~kg)(9.8 ~m ~s^ -2 )=9.8 ~N . At the top of the circle, T=F_c-W=50 ~N- 9.8 ~N= 40.2 ~N

AIIMS 2015

Exam: AIIMS
Year: 2015
Subject: Physics
Chapter: Motion in Two Dimensions

AIIMS 2015 · Physics · Motion in Two Dimensions

A string 0.5 ~m long is used to whirl a 1 ~kg stone in a vertical...

A string

0.5 \mathrm{~m}

long is used to whirl a

1 \mathrm{~kg}

stone in a vertical circle at a uniform velocity of

5 \mathrm{~m} \mathrm{~s}^{-1}

. What is the tension in the string when the stone is at the top of the circle?

Options

A $9.8 \mathrm{~N}$
B $30.4 \mathrm{~N}$
C $40.2 \mathrm{~N}$
D $59.8 \mathrm{~N}$

Correct answer

C. $40.2 \mathrm{~N}$

Step-by-step solution

The centripetal force needed to keep the stone moving at

5 \mathrm{~m} \mathrm{~s}^{-1}

is F_c=

\frac{m v^2}{r}=\frac{(1 \mathrm{~kg})\left(5 \mathrm{~ms}^{-1}\right)^2}{0.5 \mathrm{~m}}=

\mathrm{~N}

The weight of the stone is

W=m g=(1 \mathrm{~kg})\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)=9.8 \mathrm{~N}

. At the top of the circle, T=F_c-W=50

\mathrm{~N}-

9.8

\mathrm{~N}=

40.2

\mathrm{~N}

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