System of Particles and Rotational Motion

The given situation can be shown as

NB = force exerted on the ladder by the floor point B

NB = force exerted on the ladder by the floor point B

T = Tension in the rope

BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m

Mass of the weight, m = 40 kg

The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.

Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.

BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)

D is the mid-point of AB, hence we can write

AD = (1/2) x BA = 0.8 ……. (ii)

Using equation (i) and (ii), we get FE = 0.4 m

Hence, F is the mid-point of AD.

Hence G will also be the midpoint of AH.

ΔAFG and ΔADH are similar

FG = (1/2)DH = (1/2) X 0.25 = 0.125 M

In ΔADH, AH = $\frac{3AB}{4g\sin ?\theta })$ - $\surd \frac{11.46}{19.6}$ ) = $\frac{FG}{DH}=\frac{AF}{AD}=\frac{FG}{DF}=\frac{0.4}{0.8}=\frac{1}{2}$ – $\surd ({AD}^2$ ) = 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force, NC + NB = mg = 392 …… (iii)

For rotational equilibrium of the ladder, the net moment about A is

${DH}^2$ NB x BI +mg x FG + NC x CI + T x AG – T x AG = 0

$\surd \left\{\right({0.8}^2$ NB x 0.5 +40 x g + 0.125 x NC x 0.5 = 0

(NC - NB) x 0.5 = 49

NC - NB = 98 ………. (iv)

Adding equation (iii) and (iv) we get

NC = 245 N, NB = 147 N

For rotational equilibrium of the side AB, considering the moment about A

${0.25}^2$ NB x BI + mg x FG + T x AG = 0

$-$ 245 x 0.5 + 40 x 9.8 + T x AG = 0

T = 96.7 N

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