(B) Given: Mass $m = 2 \, kg$,radius $r = 1 \, m$,speed $v = 5 \, m/s$,$g = 10 \, m/s^2$.Centripetal force $F_c = \frac{mv^2}{r} = \frac{2 \times (5)^

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(B) Given: Mass $m = 2 \, kg$,radius $r = 1 \, m$,speed $v = 5 \, m/s$,$g = 10 \, m/s^2$.Centripetal force $F_c = \frac{mv^2}{r} = \frac{2 \times (5)^2}{1} = 50 \, N$.Weight $W = mg = 2 \times 10 = 20 \, N$.At the bottom of the circle,the tension $T$ is given by $T = F_c + mg = 50 + 20 = 70 \, N$.Therefore,the tension is $70 \, N$ at the bottom of the circle.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

Medium

$A$ $2 \, kg$ mass is attached to a string of length $1 \, m$ and is whirled in a vertical circle at a constant speed of $5 \, m/s$. At what position will the tension in the string be $70 \, N$? $(g = 10 \, m/s^2)$.

A
At the top of the circle
B
At the bottom of the circle
C
At the horizontal position
D
None of these

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Work, Energy, Power and Collision

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Vertical Circular Motion

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

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$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)

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In the figure shown,a particle is released from the position $A$ on a smooth track. When the particle reaches at $B$,then the normal reaction on it by the track is .........

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$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

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The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is:

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$A$ $2 \, kg$ mass is attached to a string of length $1 \, m$ and is whirled in a vertical circle at a constant speed of $5 \, m/s$. At what position will the tension in the string be $70 \, N$? $(g = 10 \, m/s^2)$.

Similar Questions

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)

In the figure shown,a particle is released from the position $A$ on a smooth track. When the particle reaches at $B$,then the normal reaction on it by the track is .........

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is:

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