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physics ncert solutions class 11th Physics Class 11th System of Particles and Rotational Motion

7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

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Vishal Baghel | Contributor-Level 10

4 months ago

Initial velocity of the cylinder, v = 5 m/s

Angle of inclination, $ω$ = 30 $\sqrt (\frac{2}{3})$

Height reached by the cylinder = h

Energy of the cylinder at point A:

KE rot + KEtrans

(1/2) I $ω$ + (1/2) m $θ$

Energy of the cylinder at point B = mgh

Using the law of conservation of energy

(1/2) I $°$ + (1/2) m $ω^{2}$ = mgh

Moment of inertia of the solid cylinder I = (1/2) mr²

Hence (1/2)(1/2)mr² $v^{2}$ + (1/2) m $ω^{2}$ = mgh

We also know v = r $v^{2}$

(1/4)m $ω^{2}$ + (1/2) m $v^{2}$ = mgh

(3/4) $ω$ = gh

h = (3/4)( $v^{2}$ = (3/4)(25/9.81) = 1.91 m

In Δ ABC, $v^{2}$ = $v^{2}$ , AB = BC / $v^{2} / g)$ = 3.82 m

Henc

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Initial velocity of the cylinder, v = 5 m/s

Angle of inclination, $ω$ = 30 $\sqrt (\frac{2}{3})$

Height reached by the cylinder = h

Energy of the cylinder at point A:

KE rot + KEtrans

(1/2) I $ω$ + (1/2) m $θ$

Energy of the cylinder at point B = mgh

Using the law of conservation of energy

(1/2) I $°$ + (1/2) m $ω^{2}$ = mgh

Moment of inertia of the solid cylinder I = (1/2) mr²

Hence (1/2)(1/2)mr² $v^{2}$ + (1/2) m $ω^{2}$ = mgh

We also know v = r $v^{2}$

(1/4)m $ω^{2}$ + (1/2) m $v^{2}$ = mgh

(3/4) $ω$ = gh

h = (3/4)( $v^{2}$ = (3/4)(25/9.81) = 1.91 m

In Δ ABC, $v^{2}$ = $v^{2}$ , AB = BC / $v^{2} / g)$ = 3.82 m

Hence the cylinder will travel 3.82 m on the inclined plane

For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to bottom is given by the relation

v = $\sin ? θ$ = $\frac{B C}{A B}$ )

for solid cylinder, $\sin ? 30 °$ = $\sqrt (\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}})$

v = $\sqrt (\frac{2 g A B \sin ? θ}{1 + \frac{k^{2}}{R^{2}}}$ ) = $k^{2}$ g AB $\frac{R^{2}}{2}$ )

The time taken to return to the bottom is :

t = $\sqrt (\frac{2 g A B \sin ? θ}{1 + \frac{1}{2}}$ = AB/ $\sqrt (\frac{4}{3}$ ( $\sin ? θ$ = $\frac{A B}{v}$ ( $\sqrt$ = $\frac{4}{3} g A B \sin ? θ)$ = 0.764 s

Therefore, the total time taken by the cylinder to return to the bottom is 2 x 0.764 = 1.53 s

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7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

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7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom?

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7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?