Mechanics> a solid sphere of mass m is situated on a...

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 .

Aakash Shinde , 9 Years ago

Grade 11

4 Answers

Vikas TU

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.

Last Activity: 9 Years ago

Karan Gupta

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

Last Activity: 7 Years ago

Suraj

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

Last Activity: 7 Years ago

Rishi Sharma

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 = 2/5mr^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 + 3F/7)/m = 10F/7m

Thanks and Regards

Last Activity: 5 Years ago