Class 11th

(a) Initial angular velocity, $\omega$ = 40 rev/min, let the final angular velocity be $\omega$

Let the moment of inertia of the boy with hands stretched be I1 and

M.I. with folded hands be I2

Given I2 = (2/5) I1

Since no external force acts on the boy, the angular momentum will remain constant.

Hence I1 $\omega$ = I2 ${\omega }_1$ ,  ${\omega }_2$ I1/I2) x ${\omega }_1$ = (5/2) x 40 = 100 rev/min

(b) Kinetic energy Ev = (1/2)I ${\omega }_2$

Hence ( Final KE / Initial KE ) = (I2 ${\omega }_2=($ )/ (I1 ${\omega }_1$ ) = { (2/5) I1 x 100 x 100 } / { I1 x 40 x 40 }

= 2.5

Mass of the cylinder, m = 20 kg

Angular speed,  $\alpha$ = 100 rad/s

Radius of the cylinder, r = 0.25 m

The moment of inertia of the solid cylinder

I = (1/2) mr² = (1/2) x 20 x (0.25)² = 0.625 m²

Kinetic energy = (1/2)I ${\omega }_s$² = (1/2) x 0.625 x (100)² = 3125 J

Angular momentum, L = I ${\omega }_h$ = 0.625 x 100 = 62.5 Js

Mass of the car, m = 1800 kg

Distance between the front and rear axles, d = 1.8 m

Distance between C.G. and the front axle = 1.05 m

Let Rf and Rb be the force exerted from ground at front and rear axles respectively.

Rf + Rb = mg = 1800 x 9.8 N = 17640 N ……. (i)

For rotational equilibrium around C.G. we have Rf x 1.05 = Rb x (1.8 – 1.05)

R_f x 1.05 = R_b x 0.75

R_f/R_b = 0.75 / 1.05

R_f = 0.71 R_b ……. (ii)

From equation (i), we get

0.71 R_b + R_b = 17640

R_b = 10316 N

R_f = 7324 N

Therefore, force exerted on each front wheel = R_f/2 =7324/2 = 3662 N

Force exerted on each rear wheel, R_b/2 = 10316 / 2= 5158 N

(i) A = {– 3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {1, 2, 3, 4, 5}

(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) x = Prime number which are divisor of 60

Factors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60

Hence, x = 2, 3, 5

?D = {2, 3, 5}

(v) E = {T, R, I, G, O, N, M, E, Y}

(vi) F = {B, E, T, R}

A free body diagram needs to be drawn.

The length of the bar, l = 2 m

T1 and T2

At translational equilibrium, we have $\stackrel{?}{L_q}=\stackrel{?}{L_R}$ = $T_1$

(T1 / T2) = ( $T_2$ / $T_1\sin ?36.9°$ = 4/3

T1 = (4/3)T2

For rotational equilibrium, on taking the torque about the centre of gravity, we have

T1 $T_2\sin ?53.1°$ x d = T2 $\sin ?53.1°$  (2-d)

T1 x 0.8d = T2 x 0.6 (2-d)

(4/3)T2 x 0.8d = T2 x 0.6 (2-d)

(4/3) x 0.8d = 0.6 (2-d)

1.07d = 1.2 – 0.6d

d = 0.72

So the c.g. of the given bar lies at 0.72 m from its left end.

2. (i) 5 $\in$ A

(ii) 8 $\notin$ A

(iii) 0 $\notin$ A

(iv) 4 $\in$ A

(v) 2 $\in$ A

(vi) 10 $\notin$ A

Let at certain instant two particles be at points P and Q, as shown in the figure.

Angular momentum of the system about point P

$\stackrel{?}{L_p}$ = mv x 0 + mv x d = mvd ……. (i)

Angular momentum of the system about point Q

$\stackrel{?}{L_q}$ = mv x d + mv x 0 = mvd ……. (ii)

Consider a point R, which is at a distance y from point Q such that QR = y

PR = d – y

Angular momentum of the system about point R

$\stackrel{?}{L_R}$  = mv x (d – y) + mv x y = mvd – mvy + mvy = mvd ……. (iii)

Comparing equations (i), (ii) and (iii) we get

$\stackrel{?}{L_p}$ = $\stackrel{?}{L_q}$ = $\stackrel{?}{L_R}$ …… (iv)

Let $\stackrel{?}{c}$ = $\stackrel{?}{b}$ ,  $\stackrel{?}{a}$ = $\stackrel{?}{a}$ ,  $\stackrel{?}{OA}$ = $\stackrel{?}{b}$

Let $\stackrel{?}{OB}$ be a unit vector perpendicular to both b and c. Hence $\stackrel{?}{c}$ and a have the same direction

Now $\stackrel{?}{OC}$ = bc $\stackrel{?}{n}$ $\stackrel{?}{n}$

= bc $\stackrel{?}{b}*\stackrel{?}{c}$ $\sin ?\theta$ = bc $\stackrel{?}{n}$

Now $\sin ?90°$ )= a. (bc $\stackrel{?}{n}$ ) = abccos $\stackrel{?}{n}$ ? = abccos0° = abc = Volume of the parallelepiped