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A solid disc of mass m is rolling with translation velocity  V m/s  .Find minimum value of V (in m/sec) to reach at height 

h = 7.5 m 

Option: 1 5

Option: 2 7

Option: 3 8

Option: 4 10

Answers (1)

best_answer

 

 

Kinetic energy of a body in combined rotation and translation -

K.E=\frac{1}{2}\, I_{cm}w^{2}+\frac{1}{2}mv_{0}\, ^{2}

- wherein

\frac{1}{2}mv_{0}\, ^{2} =  Translational kinetic energy of centre of mass

 

\frac{1}{2}\, I_{cm}w^{2}  =  Rotational kinetic energy about centre of mass

 

 Form energy conservation 

WR\Delta KE

mgh = \frac{1}{2} mv^2 + \frac{1}{2} IW^2 \\\\ mgh = \frac{1}{2} mv^2 +\frac{1}{2}[\frac{1}{2}mR^2])(\frac{V^2}{R^2})\\\\mgh = \frac{1}{2} mv^2 +\frac{1}{4}mv^2\\\\ mgh = \frac{3}{4} mv^2 \Rightarrow v = \sqrt{\frac{4}{3}gh} \\\\v= \sqrt{\frac{4}{3}\times 10\times 7.5} \Rightarrow v = 10 m/s

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Pankaj

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