Laws of Motion

Let $\theta$ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N $\cos ?\theta$ …….(1)

mr ${\omega }^2$ = N $\sin ?\theta$ ……(2)

m(R $\sin ?\theta$ ) ${\omega }^2$ = N $\sin ?\theta$

Hence N = m(R) ${\omega }^2$

Substituting the value on N in eqn (1)

mg = mR ${\omega }^2\cos ?\theta$

or $\cos ?\theta$ = g/ R ${\omega }^2$ ………(3)

As $\left|\cos ?\theta \right|$ $\le$ 1, the bead will remain at the lowermost point

g/ R ${\omega }^2$ $\le 1or\omega \le \sqrt{g}/R$

For $\omega$ = $\frac{\sqrt{2g}}{R},equation\left(3\right)$ becomes

$\cos ?\theta$ = g/ R ${\omega }^2$

$\cos ?\theta$ =(g/R)(R/2g) = ½

$\theta =60°$

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing,  $\mu$ = 0.15

Number of revs of hollow cylindrical drum = 200 rev/min = 200/60 rev/s = 3.33 rev/s

The centripetal force required is provided by the normal N of the wall on the man

N = m $v^2/R$ = m ${\omega }^2$ R

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downwards is balanced by the frictional force acting vertically upwards.

The man will not fall, if

mg $\le \mu N$

mg $\le \mu (m{\omega }^{2}R$ )

${\omega }^2\ge g/R\mu$

$g/R\mu$ = 10/ (3 $*$ 0.15)

$\omega =4.71rad/s$

When the motorcyclist is at the uppermost point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by two forces.

R + mg = m $v^2/r$ , where v is the velocity and m is the combined mass of the motorcycle and motorcyclist

Because of the balance between the forces, the motorcyclist does not fall.

The minimum speed required at the uppermost position to perform a vertical loop is given by R = 0 in the above equation.

So mg = m $v^2/r$ or v = $\sqrt{gr}$ = $\sqrt{10*25}$ = 15.8 m/s

Speed of revolution of the disc, n = $33\frac{1}{3}$ rev/min = 100/3 rpm = 100/ (3 $*60)=0.56rps$

Angular acceleration $\omega$ = 2 $\pi n$ = 2 $*\frac{22}{7}*0.56$ = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force $m{v}^2/r\le \mu mg$ …. (1)

As v = r $\omega$ , the above equation becomes mr ${\omega }^2/r\le \mu mg$

r $\le \mu g/{\omega }^2$

r $\le$  (0.15 $*10)/{3.492}^2$ = 12 cm

For coin A, r = 4 cm

The condition (r $\le$ 12 ) is satisfied for the coin placed at r = 4 cm, so coin A will revolve with the disc.

The condition (r $\le$ 12 ) is not satisfied for the coin placed at r = 14 cm, so coin B will not revolve with the disc.

Force on the box, F = MA = 40 $*$ 2 N = 80 N

Frictional force, Ff = $?mg$ = 0.15 $*40*10=$ 60 N

Net force = F – Ff = 80 – 60 = 20 N

From the equation F = ma, we get the backward acceleration produced in the box

a = 20/40 = 0.5 m/s2

From the equation s = ut + $\frac{1}{2}a{t}^2$ , to travel s = 5 m by the box to fall off from the truck, we get

5 = 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$

5 = 0.25 $t^2$ , t = 4.47 s

The travel of truck during t = 4.47 s is

= 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$ = 0.5 $*2*{4.47}^2$ = 19.98 m

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley  $\mu$ = 0.18

Acceleration of the trolley = 0.5 m/s2

(a) Force experienced by block, F = MA = 15 $*$ 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

Force of friction, Ff = $\mu mg$ = 0.18 $*15*10$ N = 27 N

Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

(b) When an observer moves with the trolley, the trolley will appear to be at rest

Mass of the block = 25 kg

Mass of the man = 50 kg

Acceleration due to gravity = 10 m/s2

Weight of the block = 250 N

Weight of the man = 500 N

In the 1st case, the man lifts the block directly, applying an upward force, same as the block = 250 N

Due to Newton's 3rd law of motion, the downward reaction of the man on the floor = 500 N + 250 N = 750 N

In the 2nd case, the man applies a downward force of 25 kg wt. According to Newton's 3rd law, the action on the floor by the man 500N-250N = 250N

The man should adopt the second case

Given

Radius, r = 30m

Speed, v = 54 km/h = 15 m/s

The mass of the train, m = 106 kg

(a) The required centripetal force is provided by the rails to the wheels of the train

(b) The angle of banking required to prevent wearing out of the rails

$\tan ?\theta$ = $v^2$ / rg = ${15}^2$ / (30 $*10)$

$\theta$ = 37 $°$