Question

Question number 1 &2 on
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Answer 1

Answer on Question#56544 - Physics - Mechanics - Relativity

practice Exercise 5.5

(i) A solid wooden door $1\mathrm{m}$ wide and $2\mathrm{m}$ high is hinged along one side and has a total mass of $50\mathrm{kg}$ . Initially open and at rest, the door is struck at its centre by a handful of sticky mud of mass $0.5\mathrm{kg}$ . Travelling $12\mathrm{m/s}$ just before impact. Find the final angular velocity of the door.

$$\left[ \frac {r _ {1}}{1 0 3} m d / s \right]$$

(ii) A thin uniform square plate with side $l$ and mass $M$ can rotate freely about a stationary vertical axis omitting with one of its sides. A small ball of mass $m$ flying with velocity $v$ at right angles to the plate strikes elastically the centre of it. Find the velocity of the ball $v'$ after the impact and the horizontal component of the force which the axis will exert on the plate after the impact.

$$\left[ \frac {v ^ {\prime} - 1 M}{m s - 1 0 3} v \right]$$

(iii) A small disc of mass $m$ slides down a smooth hill of height $h$ without initial velocity and gets also a plank of mass $M$ lying on the horizontal edge on the base of the hill as shown in Figure 1. Find the fixation between the disc and

Solution:

(i) The angular momentum of the mud is

$$L = m v r,$$

Where $v = 12\frac{\mathrm{m}}{\mathrm{s}}$ , $m = 0.5\mathrm{kg}$ and $r = \frac{1}{2}\mathrm{m}$ - is the distance from the axis of rotation to the door's centre.

The moment of inertia of the door is

$$I _ {d} = \frac {1}{3} M l ^ {2},$$

Where $M = 50\mathrm{kg}$ and $l = 1\mathrm{m}$ .

The moment of inertia of the mud (about the axis of rotation) is

$$I _ {m} = m r ^ {2}$$

Since the angular momentum is conserved the final angular momentum $(I_d + I_m)\omega$ must be equal to the initial $L$ :

$$m v r = \left(I _ {d} + I _ {m}\right) \omega$$

Thus

$$\omega = \frac {m v r}{I _ {d} + I _ {m}} = \frac {m v r}{\frac {1}{3} M l ^ {2} + m r ^ {2}} = \frac {0 . 5 \mathrm {k g} \cdot 1 2 \frac {\mathrm {m}}{\mathrm {s}} \cdot 0 . 5 \mathrm {m}}{\frac {1}{3} 5 0 \mathrm {k g} \cdot (1 \mathrm {m}) ^ {2} + 0 . 5 \mathrm {k g} \cdot (0 . 5 \mathrm {m}) ^ {2}} = \frac {7 2}{4 0 3} \frac {\mathrm {r a d}}{\mathrm {s}}$$

(ii) According to the law of conservation of angular momentum

$$\frac {1}{3} M l ^ {2} \omega + m v ^ {\prime} \frac {l}{2} = m v \frac {l}{2}$$

According to the law of conservation of energy

$$\frac {\frac {1}{3} M l ^ {2} \omega^ {2}}{2} + \frac {m v ^ {\prime 2}}{2} = \frac {m v ^ {2}}{2}$$

(1):

$$v - v ^ {\prime} = \frac {2}{3} \frac {M}{m} l \omega$$ $$(v - v ^ {\prime}) ^ {2} = \left(\frac {2}{3} \frac {M}{m} l \omega\right) ^ {2}$$

(2):

$$v ^ {2} - v ^ {\prime 2} = \frac {1}{3} \frac {M}{m} l ^ {2} \omega^ {2}$$

Dividing (3) by (4) we obtain

$$\frac {(v - v ^ {\prime}) ^ {2}}{v ^ {2} - v ^ {\prime 2}} = \frac {\left(\frac {2}{3} \frac {M}{m} l \omega\right) ^ {2}}{\frac {1}{3} \frac {M}{m} l ^ {2} \omega^ {2}}$$ $$\frac {v - v ^ {\prime}}{v + v ^ {\prime}} = \frac {4}{3} \frac {M}{m}$$ $$v ^ {\prime} = \frac {3 m - 4 M}{4 M + 3 m} v$$

Thus the angular velocity $\omega$ is

$$\omega = \frac {3 m}{2 M} \frac {1}{l} (v - v ^ {\prime}) = \frac {v}{l} \frac {1 2 m}{4 M + 3 m}$$

The horizontal component of the force acting on the axis is given by the centrifugal force of the plate which is (integral over all vertical stripes of width $dr$ at distance $r$ from the axis of revolution)

$$F = \int_ {0} ^ {l} M \omega^ {2} \frac {r}{l} d r = \frac {1}{2} M \omega^ {2} l = \frac {1}{2} M \left(\frac {v}{l} \frac {1 2 m}{4 M + 3 m}\right) ^ {2} l = 7 2 M \frac {v ^ {2}}{l} \frac {m ^ {2}}{(4 M + 3 m) ^ {2}}$$

Answer:

(i) $\frac{72\mathrm{rad}}{403\mathrm{s}}$

(ii) $v^{\prime} = \frac{3m - 4M}{4M + 3m} v$

$$F = 7 2 M \frac {v ^ {2}}{l} \frac {m ^ {2}}{(4 M + 3 m) ^ {2}}$$

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