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Grade: 11

                        

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 .

4 years ago

Answers : (4)

Vikas TU
13786 Points
							
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.
4 years ago
Karan Gupta
51 Points
							
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
2 years ago
Suraj
11 Points
							
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
one year ago
Rishi Sharma
askIITians Faculty
614 Points
							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
4 months ago
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