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  • Please help! A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular accelerat

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (neglect the mass and thickness of the string):

 

  • Option 1)

    20 rad/s2

  • Option 2)

    16 rad/s2

  • Option 3)

    12 rad/s2

  • Option 4)

    10rad/s2

Answers (1)

best_answer

 

Analogue of second law of motion for pure rotation -

\vec{\tau }=I\, \alpha

- wherein

Torque equation can be applied only about two point

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

 

Draw FBD of hollow cylinder

It is rolling without slipping.

So point of contact of ground & hollow cylinder will be at rest.

So a=R\alpha \cdots (1)

apply \sum F_{x}=ma_{x}

\Rightarrow 40+f=ma=mR{\alpha }\cdots (2)

apply \tau =I\alpha \: \: \: (about\: \: 0\: \: \: point)

40\times R-f\times R=mR^{2}\alpha \cdots \left ( 3 \right )

From (2) & (3)

\alpha =\frac{40}{MR}=16rad/s^{2}


Option 1)

20 rad/s2

Option 2)

16 rad/s2

Option 3)

12 rad/s2

Option 4)

10rad/s2

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