System of Particles and Rotational Motion

This is a multiple choice type question as classified in NCERT Exemplar

(b) When the small piece Q is removed and glued to the centre of the plate . the mass closer to the z axis . hence moment of inertia decreases.

This is a multiple choice type question as classified in NCERT Exemplar

(d) we know that angular acceleration $\alpha =\frac{dw}{dt}$ , given w=constant

where w is the angular velocity of the disc

$\alpha =\frac{dw}{dt}=\frac{0}{dt}=0$

Hence angular acceleration is zero.

This is a short answer type question as classified in NCERT Exemplar

Moment of force $\tau$ ₁= F (a)….anticlockwise

Moment of weight mg $\tau$ ₂=mg (a/2)……. (clockwise)

Cube will not exhibit motion then $\tau$ _{1 $=\tau$ 2}

$F*a=mg*\frac{a}{2}$

F=mg/2

Cube will rotate when $\tau$ _{1 $>\tau$ 2}

$F*a>mg*\frac{a}{2}$

F>mg/2

At a/3 from point A then

mg $*\frac{a}{3}=F*aorF=\frac{mg}{3}$

when F=mg/4 which is less then mg/3 there will be no motion.

This is a short answer type question as classified in NCERT Exemplar

The moment of inertia of the object I = $\sum m$ r² [sum of moment of inertia of each constituents particles]

All the mass in a cylinder lies at a distance R from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than R.

This is a long answer type question as classified in NCERT Exemplar

Frictional force f is acting in the opposite direction of F . let the acceleration of centre of mass of disc be a then

F-f=Ma where M is the mass of the disc

$\alpha =\frac{a}{R}$

$\tau =I\alpha$

fR= (1/2 MR²) $\alpha$

so fR= (1/2MR²) (a/R)

Ma=2f

From the above equation

F = F/3

F< $\mu N$ = $\mu Mg$

f/3 < $\mu Mg$

F=3 $\mu Mg$

This is a long answer type question as classified in NCERT Exemplar

(a) Situation as given below

(b) F' =F=F'' where F' and F'' are external forces through support.

So F_net=O

External torque = F $*3R$ (anticlockwise)

(c) Let w₁ and w₂ be the final angular velocities of smaller and bigger drum in both clockwise and anticlockwise . finally there will be no friction

hence Rw₁=2Rw₂

so $\frac{w_1}{w_2}=2$

This is a long answer type question as classified in NCERT Exemplar

(a) Before being bought in contact with the table the disc was in pure rotational motion. Hence V_cm=0

(b) when the disc is placed in contact with table due to friction centre of mass acquires some linear velocity.

(c) when the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.

(d) friction is responsible for the effects ib b and c

(e) when rolling starts V_cm=wR

(f) time period for rolling to begin is t= $\frac{R{w}_o}{\mu g(1+\frac{m{R}^2}{I})}$

This is a long answer type question as classified in NCERT Exemplar

(a) yes the law of conservation of angular momentum can be applied because there is no net external torque on the system of the two discs.

External forces gravitation and normal reaction act through the axis of rotation, hence produce no torque.

(b) by conservation of angular momentum

L_f= L_i

Iw=I₁w₁+I₂w₂

So w= $\frac{I_1{w}_1+I_2{w}_2}{I_1+I_2}$

(c) K_f= $\frac{1}{2}(I_1+I_2)\frac{{(I_1{w}_1+I_2{w}_{2)}}^2}{I_1+I_2}$

K_f= ½ (I₁w₁²+I₂w₂²)

$?k=$ K_f-K_i =- $\frac{-I_1{I}_2}{2\left(I_1+I_2\right)}(w_1-w_2)$ ²<0

(d) hence there is loss of KE of the system. The loss of kinetic energy is mainly due to work against the friction.