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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 .

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 .

Grade:11

4 Answers

Vikas TU
14149 Points
7 years ago
Torque from the centre of Rotation we get,
T = I*a
a is angular accln.
F*R = ((2/5)MR^2 + MR^2)a
we get a = 5F/7MR rad/sec.
Now accln. = a/R
Hence Acom = 5F/7M
where F is the tangential constant Force And M is th mass of the sphere.
Karan Gupta
51 Points
5 years ago
FR – fR = (2/5)MR^2*a/R
F – f = (2/5)Ma....................(1)
F + f = Ma...........................(2)
 
Adding, 2F = (7/5)Ma
a = 10F/7m ….......................Answer
Suraj
11 Points
5 years ago
Torque from the centre of Rotation we get,
T = I*a
a is angular accln.
F*R = ((2/5)MR^2 + MR^2)a
we get a = 5F/7MR rad/sec.
Now accln. = a/R
Hence Acom = 5F/7M
where F is the tangential constant Force And M is th mass of the sphere.    Ok
Rishi Sharma
askIITians Faculty 646 Points
3 years 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^2
Using: 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 zero
i.e. a=rα
∴ (F+f)/m ​= r5(F−f)/2mr​
Or, 2F + 2f = 5F−5f
⟹ f = 3F​/7
Thus from (2), we get: a = (F + 3​F/7)/m ​= 10F/7m​

Thanks and Regards

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