Class 11th

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

Angle of inclination, $\omega$ = 30 $\surd \left(\frac{2}{3}\right)$

Height reached by the cylinder = h

Energy of the cylinder at point A:

KE rot + KEtrans

(1/2) I $\omega$ + (1/2) m $\theta$

Energy of the cylinder at point B = mgh

Using the law of conservation of energy

(1/2) I $°$ + (1/2) m ${\omega }^2$ = mgh

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

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

We also know v = r $v^2$

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

(3/4) $\omega$ = 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 = $\sin ?\theta$ = $\frac{BC}{AB}$ )

for solid cylinder, $\sin ?30°$ = $\surd \left(\frac{2gh}{1+\frac{k^2}{R^2}}\right)$

v = $\surd (\frac{2gAB\sin ?\theta }{1+\frac{k^2}{R^2}}$ ) = $k^2$ g AB $\frac{R^2}{2}$ )

The time taken to return to the bottom is :

t = $\surd (\frac{2gAB\sin ?\theta }{1+\frac{1}{2}}$ = AB/ $\surd (\frac{4}{3}$ ( $\sin ?\theta$ = $\frac{AB}{v}$ ( $\surd$ = $\frac{4}{3}gAB\sin ?\theta )$ = 0.764 s

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

Height of the plane = h

Velocity of the sphere at the bottom of the plane = v

At the top of the plane, the total energy of the sphere = potential energy = mgh

At the bottom of the plane, the sphere has both translational and rotational energies.

Hence, total energy = (1/2)mv² + (1/2)I $\frac{-MR}{8}$

Using the law of conservation of energy, we can write: (1/2)mv²+ (1/2)I $\frac{4}{3M}$ = mgh …(1)

For a solid sphere, the moment of inertia, I = (2/5)mr²

The equation (1) becomes (1/2)mv² + (1/2)( (2/5)mr² $-\frac{R}{6}$ = mgh

(1/2) v² + (1/5)r² ${\omega }^2$ = gh

From the relation v = ${\omega }^2$ , we get

(1/2) v²+ (1/5) v²= gh

v = ${\omega }^2$

Since v depends only on g (constant) and h (height = constant), the velocity at the bottom remains same from whichever plane the sphere rolled.

(b) Yes

(c)

Consider two inclined planes with inclinations ${\omega }^2$ and $r\omega$ related as $\surd \left(\frac{10}{7}\right)gh$ > ${\theta }_1$ .

The acceleration produced in the sphere where it rolls down the plane inclined at ${\theta }_2$ is given as g sin ${\theta }_2$ . Similarly for the plane with inclination ${\theta }_1$ is given as g sin ${\theta }_1$

Let R1 and R2 be the normal reaction to the sphere on these two planes

Since ${\theta }_1$ > ${\theta }_2$ , ${\theta }_2$ > sin ${\theta }_2$ ….(i)

${\theta }_1$ > ${sin\theta }_2$ …….(ii)

From the equation v = u +at where u = 0 and v = constant, we get t ${\theta }_1$ (1/a)

For inclination $a_2$ ,t₁ $a_1$ (1/a₁) and t₂$\alpha$ (1/a₂)

Therefore t₂ < t₁

So the sphere will take a longer time to reach plane with smaller inclination.

Radius of the original disc = R

Mass of the smaller disc = $\tau$ = (1/4) $\omega$ = M/4

Let O and O' be the respective centers of the original disc and the cut out disc respectively. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O'.

It is given, OO' = R/2

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are

M – concentrated at O and M/4 concentrated at O'

Let x be the distance through which the centers of masses of two masses shifts from O.

The relation between the centers of masses is given as

x = (m₁r₁ + m₂r₂)/ (m₁ +m₂)

= $\pi R^2\sigma$ = $\pi {\left(\frac{R}{2}\right)}^2\sigma$ = $\pi R^2\sigma$ x $\frac{Mx0-M^{\text{'}}x\left(\frac{R}{2}\right)}{M+(-M^{\text{'}})}$ = $\frac{-\left(\frac{M}{4}\right)x\left(\frac{R}{2}\right)}{M-M/4}$