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```
a solid sphere of mass m is situated on a horizontal surface and a tangential force act at the top of the sphere. if the sphere rolls without slipping then the accelaration of the centre of sphere would be – . pls give the answer and explain it in steps .

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4 years ago

```							Torque from the centre of Rotation we get,T = I*aa is angular accln.F*R = ((2/5)MR^2 + MR^2)awe get a = 5F/7MR rad/sec.Now accln. = a/RHence Acom = 5F/7Mwhere F is the tangential constant Force And M is th mass of the sphere.
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4 years ago
```							FR – fR = (2/5)MR^2*a/RF – f = (2/5)Ma....................(1)F + f = Ma...........................(2) Adding, 2F = (7/5)Maa = 10F/7m ….......................Answer
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2 years ago
```							Torque from the centre of Rotation we get,T = I*aa is angular accln.F*R = ((2/5)MR^2 + MR^2)awe get a = 5F/7MR rad/sec.Now accln. = a/RHence Acom = 5F/7Mwhere F is the tangential constant Force And M is th mass of the sphere.    Ok
```
one year ago
```							Dear Student,Please find below the solution to your problem.Let the angular acceleration of the sphere be α.Moment of inertia of the sphere, I=2​/5mr^2Using: τ = Iα∴  (F−f)r = 2/5​mr^2α     ⟹ α = 5(F−f)/2mr​     ............(1)Acceleration of its centre, a = (F+f)/m​      ......(2)As the sphere performs pure rolling, thus acceleration of the point O is zeroi.e. a=rα∴ (F+f)/m ​= r5(F−f)/2mr​Or,  2F + 2f = 5F−5f⟹ f = 3F​/7Thus from (2), we get: a = (F + 3​F/7)/m ​= 10F/7m​Thanks and Regards
```
4 months ago
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