A Solid Disc and a Ring, Both of Radius 10 cm Are Placed on a Horizontal Table Simultaneously, with Initial Angular Speed Equal to 10 π Rad S-1. Which of the Two Will Start to Roll Earlier? the Coefficient of Kinetic Friction Is μK = 0.2 - Physics

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is μk = 0.2.

Solution 1

Disc Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω=10 π rad s–1

Coefficient of kinetic friction, μk = 0.2

Initial velocity of both the objects, u = 0

Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma

μkmg= ma

Where,

a = Acceleration produced in the objects

m = Mass

∴a = μkg … (i)

As per the first equation of motion, the final velocity of the objects can be obtained as:

v = u + at

= 0 + μkgt

= μkgt … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause the reduction in the initial angular speed.

Torque, τ= –Iα

α = Angular acceleration

μxmgr = –Iα

:.alpha = (-mu_kmgr)/I ....(iii)

Using the first equation of rotational motion to obtain the final angular speed:

omega = omega_0 + alphat

= omega_0 + (-mu_kmgr)/It  ....(iv)

Rolling starts when linear velocity, v = rω

:.v= r(omega_0 - (mu_kgmrt)/I) ...(v)

Equating equations (ii) and (v), we get:

mu_kgt = r (omega_0 - (mu_kgmrt)/I)

= romega_0 - (mu_kgmr^2t)/I   .....(vi)

For the ring: I =mr^2

:. mu"gt" = romega_0 - (mu_kgmr^2t)/(mr^2)

= romega_0 -mu_kgmt_r

2mu_k"gt" = romega_0

:. t_r = (rw_0)/(2mu_kg)

= (0.1xx10xx3.14)/(2xx0.2xx9.8) = 0.80s .....(vii)

For the disc: I = 1/2mr^2

:.mu_k"gt"_d = rw_0 (mukgmr^2t)/(1/2mr^2)

= romega_0 - 2mu_k"gt"

3mu_k"gt"_d = romega_0

:.t_d = (romega_0)/(3mu_kg)

= (0.1xx10xx3.14)/(3xx0.2xx9.8) = 0.53 s  ...(viii)

Since t_d > t_r the disc will start rolling before the ring.

Solution 2

When a disc or ring starts rotatory motion on a horizontal surface, an initial translational velocity of the centre of mass is zero.

The frictional force causes the centre of mass to accelerate linearly but frictional torque causes angular retardation. As force of normal reaction N = mg, hence frictional force f = uk N = uk mg.
For linear motion f = uk . mg = ma ———-(i)

and for rotational motion t = f.R = mu_k mg.R = -ialpha ....(ii)

Let perfect rolling motion starts at time t, when v =  Romega

From (i) a =  mu_k.g

v = u + at = 0 + mu_k.g.t

From (ii) alpha  = - (mu_k.mgR)/I = -( mu_k.mgR)/(mK^2) = - (mu_k.gR)/K^2  ....(iii)

:. omega = omega_0 + alphat = omega_0 - (mu_k.gR)/K^2 t  ...(iv)

Since v  =  Romega =, "hence " mu_k.g.t = R[omega_0 - mu_k. (gR)/K^2t]

=> t^2 = (Romega_0)/(mu_k .g(1+R^2/K^2))

For disc K^2 =  R^2/2, "hence " t = (omega_0R)/(mu_k.g(1+R^2/(R^2"/"2))) = (omega_0R)/(3mu_k.g)

For ring K^2 = R^2 hence  t = (mu_0R) /(mu_k.g(1+R^2/R^2)) = (omega_0R)/(2mu_k.g)

Thus it is clear that disc will start to roll earlier. The actual time at which disc start rolling will be

t = (omega_0R)/(2mu_k.g) = ((10pi)xx(0.1))/(3xx(0.2)xx9.8) = 0.538

Is there an error in this question or solution?

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NCERT Class 11 Physics Textbook
Chapter 7 System of Particles and Rotational Motion
Q 30 | Page 181