Laws of Motion

• Question 565

## A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The frictional forceconverts translational energy to rotational energy dissipates energy as heat decreases the rotational motion decreases the rotational and translational motion

Solution

A.

converts translational energy to rotational energy

When a body rolls down without slipping along an inclined plane of inclination θ, it rotates about a horizontal axis through its centre of mass and also its centre of mass moves. Therefore, rolling motion may be regarded as a rotational motion about an axis through its centre of mass plus a translational motion of the centre of mass. As it rolls down, it suffers loss in gravitational potential energy provided translational energy due to frictional force is converted into rotational energy.

Question 566

## A block B is pushed momentarily along a horizontal surface with an initial velocity v. If µ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time$\frac{\mathrm{\nu }}{\mathrm{g\mu }}$ $\frac{\mathrm{g\mu }}{\mathrm{\nu }}$ $\frac{\mathrm{g}}{\mathrm{\nu }}$ $\frac{\mathrm{\nu }}{\mathrm{g}}$

Solution

A.

$\frac{\mathrm{\nu }}{\mathrm{g\mu }}$

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
force applied  =  frictional force

μmg = m$\mathrm{\alpha }$

where μ - constant of proportionality

Question 567

## Diwali rocket is ejecting 50 g of gases/s at a velocity of 400 m/s. The accelerating force on the rocket will be22 dyne 20 N 20 dyne 100 N

Solution

B.

20 N

The accelerating force on the rocket

= upward thrust = $\frac{∆\mathrm{m}}{∆\mathrm{t}}.\mathrm{u}$

Given:-

,

ν  = 400 m/s

Accelerating force = 50 × 10 ×  400

= 20 N

Question 568

## A weightless thread can bear tension upto 37 N. A stone of mass 500 g is tied to it and revolved in a circular path of radius 4 min a vertical plane. If g = 10 ms-2,  then the maximum angular velocity the stone will be2 rad s-1 4 rad s-1 8 rad s-1 16 rad s-1

Solution

B.

According to second law,  the force f provding the acceleration is

F = $\frac{{\mathrm{mv}}^{2}}{\mathrm{R}}$

where m is the mass of the body. This force directed towards the centre is called the centripetal force. For a stone rotated in a circle by a string, the centripetal force is provided by the string.

Maximum tension in the thread is given by

Tmax = mg + $\frac{{\mathrm{mv}}^{2}}{\mathrm{r}}$

⇒         Tmax = mg + mrω2               (  )

ω2

Given

Tmax = 37 N,   m = 500 g = 0.5 kg,

g = 10 ms-2,    r = 4 m

∴

⇒              ω2 = 16

⇒              ω = 4 rad s-1