Question

A spherical ball of mass m and radius r rolls without slipping on rough concave surface of large radius r.It makes small oscillations about lowest point.Find time period

Answer 1

Answer:

A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period. By energy method total energy in SHM is constant

Answer 2

Answer:

Explanation:

Given, mass of the ball = m, and radius of the spherical ball = r.

Moment of Inertial of the Sphere about the axis passing through its centre =  2/5 mr².

''Now, moment of Inertial of the spherical ball about the axis passing through its surface parallel to the axis through the centre will be calculated by the parallel axis theorem.''

Thus, Moment of inertial about the tangent to sphere as an axis = 2/5 mr² + mr²

= 7/5 mr²

Now, Let the linear displacement of the spherical drop from the mean position be x.  

''Refer to the attachment for the understanding.''

Restoring force = mgSinθ.

Normal force = mgcosθ.

Now, the motion is little angular also. Thus,

 Torque about the point of contact = mgrSinθ

Now, τ = Iα

∴ α = τ/T

α = 5mgrSinθ/7mr²

α = 5gSinθ/7r

Now, Since, Oscillations are small, thus θ is small.

∴ Sinθ ≈ θ

∴ α = 5gθ/7r

  

Now, we know that a = rα

∴ a = 5gθ/7

Now, θ =x/(R-r)    [From Geometry.]

∴ a = 5gx/7(R-r) ={5g/7(R-r)}x  

Hence the acceleration is proportional to the displacement thus motion is simple harmonic motion with ⍵² = 5g/7(R-r)}

∴  ω ⇒       $2\pi \sqrt{\frac{5g}{7(R-r} }$

Also,  T = $\frac{2\pi }{w}$

∴  T = $2\pi \sqrt{\frac{7(R-r)}{5g} }$

Hence, the time period of the given S.H.M. is   T = $2\pi \sqrt{\frac{7(R-r)}{5g} }$

Hope it helps.