Question 461

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Solution

We have,

Mass of the stone,*m* = 0.25 kg

Radius of the circle, Mass of the stone,

Number of revolution per second, n =

40 / 60 = 2 / 3 rps

Angular velocity, ω = = 2πn

The tension in the string provides the centripetal force.

i.e.,

T = F

=

=

=

= 0.25 × 1.5 × (2 × 3.14 × ()

= 6.57 N

Maximum tension in the string,

∴ v

= (200 × 1.5 / 0.25)

= (1200)

= 34.64 m/s

Therefore, the maximum speed of the stone is 34.64 m/s.

Question 462

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Solution

(b) The stone flies off tangentially from the instant the string breaks.

When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Question 463

Solution

Between x = 0 and x = 2 cm, ball will be rebounding between two walls.

After every 2 s,

Magnitude of the impulse received by the ball = 0.08 × 10^{–2} kg m/s

The given graph shows that a body changes its direction of motion after every 2 s.

Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions*x* = 0 and *x* = 2 cm.

The slope of the position-time graph reverses after every 2 s, the ball collides with a wall after every 2 s.

Therefore, ball receives an impulse after every 2 s.

Given,

Mass of the ball,*m* = 0.04 kg

The slope of the graph gives the velocity of the ball.

Using the graph,

Initial velocity, (*u*) =

Velocity of the ball before collision,*u* = 10^{–2} m/s

Velocity of the ball after collision,*v* = –10^{–2} m/s

Here, the negative sign arises as the ball reverses its direction of motion.

Magnitude of impulse = Change in momentum

= | mv -*mu* |

= | 0.04 (*v* - *u*) |

= | 0.04 (-10^{-2} - 10^{-2}) |

= 0.08 × 10^{-2} kg m/s

After every 2 s,

Magnitude of the impulse received by the ball = 0.08 × 10

The given graph shows that a body changes its direction of motion after every 2 s.

Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions

The slope of the position-time graph reverses after every 2 s, the ball collides with a wall after every 2 s.

Therefore, ball receives an impulse after every 2 s.

Given,

Mass of the ball,

The slope of the graph gives the velocity of the ball.

Using the graph,

Initial velocity, (

Velocity of the ball before collision,

Velocity of the ball after collision,

Here, the negative sign arises as the ball reverses its direction of motion.

Magnitude of impulse = Change in momentum

= | mv -

= | 0.04 (

= | 0.04 (-10

= 0.08 × 10

Question 464

Lowest Point | Highest Point | |

a | mg – T_{1} |
mg + T_{2} |

b | mg + T_{1} |
mg – T_{2} |

c | mg + T_{1} – (mv_{1}^{2}) / R |
mg – T_{2} + (mv_{1}^{2}) / R |

d | mg – T_{1} – (mv_{1}^{2}) / R |
mg + T_{2} + (mv_{1}^{2}) / R |

T

T_{2} and v_{2} denote corresponding values at the highest point.

Solution

The free body diagram of the stone at the lowest point is shown in the figure below:

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force.

i.e.,*F*_{net} = *T* - *mg* = ...(i)

where,

v_{1} is the velocity at the lowest point.

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion,

* T* + *mg* = ...(ii)

where, v_{2 }is the velocity at the highest point.

From equations (i) and (ii),

Net force acting at the lowest = (T - mg)

Net force at the highest points = (*T* + *m*g)

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force.

i.e.,

where,

v

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion,

where, v

From equations (i) and (ii),

Net force acting at the lowest = (T - mg)

Net force at the highest points = (