The free body diagram of the above figure is:

Given,

Length of the plank, l = 2 m

θ_{1} = 35^{0} and θ_{2 }=55^{0}

Let T_{1} and T_{2} be the tensions produced in the left and right strings respectively.

So at translational equilibrium we have;

T_{1}sinθ_{1} = T_{2}sinθ_{2}

\( \dfrac{T_1}{T_2}=\dfrac{sinθ_2}{sinθ_1}=\dfrac{ sin55}{ sin35}\)

\( \dfrac{T_1}{T_2}=\dfrac{0.819}{0.573 }\)= 1.42

T_{1} = 1.42T_{2}

_{ } Let ‘d’ be the distance of the center of gravity of the plank from the left.

For rotational equilibrium about the centre of gravity :

T_{1}cos35 x d = T_{2 }cos 55 (2 – d)

(\( \dfrac{T_1}{T_2}\))x0.82d = (2 x 0.57 – 0.57d)

Substituting T_{1} = 1.42T_{2}

1.42 x 0.82 d + 0.57 d = 1.14

1.73d = 1.14

Therefore d = 0.65m

Answered by Abhisek | 1 year agoSeparation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

**(i)** Show p = p’_{i} + m_{i}V

Where p_{i} is the momentum of the i^{th} particle (of mass m_{i} ) and p’_{i} = m_{i}v_{i}‘. Note v’_{i} is the velocity of the i^{th} particle with respect to the centre of mass.

Also, verify using the definition of the centre of mass that Σp’_{i} = 0

**(ii)** Prove that K = K′ + \( \dfrac{1}{2}\)MV^{2}

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken relative to the centre of mass and MV^{2} /2 is the kinetic energy of the translation of the system as a whole.

**(iii)** Show L = L’+ R × MV

where L’ = ∑r’_{i} × p’_{i} is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note r’_{i} = r_{i} – R, rest of the notation is the standard notation used in the lesson. Note L’ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

**(iv)** Prove that :

\( \dfrac{dL’}{dt}\)= ∑ r’_{i} x \( \dfrac{dp’}{dt}\)

Further prove that :

\( \dfrac{dL’}{dt}\) = τ’_{ext}

Where τ’_{ext} is the sum of all external torques acting on the system about the centre of mass. ( Clue : Apply Newton’s Third Law and the definition of centre of mass . Consider that internal forces between any two particles act along the line connecting the particles.)

Read each statement below carefully, and state, with reasons, if it is true or false;

**(a)** During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

**(b) **The instantaneous speed of the point of contact during rolling is zero.

**(c)** The instantaneous acceleration of the point of contact during rolling is zero.

**(d) **For perfect rolling motion, work done against friction is zero.

**(e)** A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µ_{s} = 0.25.

**(a)** How much is the force of friction acting on the cylinder?

**(b)** What is the work done against friction during rolling?

**(c)** If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with an 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 µ_{k} = 0.2.

Explain why friction is necessary to make the disc in Figure roll in the direction indicated.

**(a)** Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

**(b)** What is the force of friction after perfect rolling begins?