Get Answers to all your Questions

header-bg qa

A solid cylinder of mass 50 Kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s-2 is :

  • Option 1)

    25 N

  • Option 2)

    50 N

  • Option 3)

    78.5 N

  • Option 4)

    157 N

 

Answers (1)

best_answer

As learnt in @7077

m=50kg, r=0.5m

\alpha =2 rev/s^{2}

Torque by Tension in string

T\times R= T\times 0.5=\frac{T}{2} N-M

T=I\alpha \Rightarrow \frac{T}{2}=I\alpha

\Rightarrow \frac{T}{2}=\left ( \frac{MR^{2}}{2} \right )\times (2\times 2\pi )

\frac{T}{2}=\frac{50\times (0.5)^{2}\times 4\pi }{2}

T=157N


Option 1)

25 N

This solution is incorrect.

Option 2)

50 N

This solution is incorrect.

Option 3)

78.5 N

This solution is incorrect.

Option 4)

157 N

This solution is correct.

Posted by

Aadil

View full answer