Chapter 6, Problem 1OQ

(a)

To determine

Rank of the acceleration for the cases.

Answer to Problem 1OQ

Rank of the acceleration for the cases is $A>C=D>B=E=0$.

Explanation of Solution

The boy moves with constant velocity around the whole track. The passes through various regions in which some are straight and some are semi circular in shape.

The moment boy reaches circular path it expereinces acceleration force due to centripetal force.

Write the expression for the acceleration of boy.

    $a_c=\frac{v^2}{r}$                                                                                                          (I)

Here, $a_c$ is the acceleration of the boy on circular path, $v$ is the velocity of boy and $r$ is the radius of circular path.

Write the acceleration of the boy on straight path.

    $a=\frac{dv}{dt}$                                                                                                         (II)

Here, $a$ is the acceleration of the boy on straight path, $\frac{dv}{dt}$ is the rate of change of velocity with time.

The moment boy reaches the straight path and moves with constant velocity.

Substitute $0\text{m/s}$ for $\frac{dv}{dt}$ in above equation.

    $a=0$                                                                                                           (III)

Therefore, acceleration is zero.

Conclusion:

Case (A)

The moment boy reaches point A, the radius of  semicircular path is small as compared to point C.

Acceleration is inversely proportional to radius of  circular path from equation (I).

Therefore acceleration increases as radius of circular path decreases.

Case (B)

Point B lies on the straight path. On this straight path boy moves with constant velocity.

The velocity does not change with time therefore acceleration becomes zero from equation (III).

Case (C)

Point C lies on the circular path.. This path has large radius as compared to point (A).

The moment boy reaches circular path it experiences acceleration The acceleration is inversely proportional to radius from eqaution (I).

Therefore the acceleration decreases as the radius inscreases.

Case (D)

Point D also lies on the circular path of larger radius . Therefore, boy experiences acceleration force.

The radius of the path is large, therefore acceleration is small from equation (I).

Case (E)

Point E is a straight path. The velocity of boy remains constant with time.

Therefore acceleration becomes zero from equation (III).

Thus, Rank of the acceleration for the cases is $A>C=D>B=E=0$.

(b)

To determine

The direction of velocity of boy at points A,B and C.

Answer to Problem 1OQ

The direction of velocity of boy at points A,B and C is $\text{North}$, $\text{West}$ and $\text{South}$ respectively.

Explanation of Solution

Direction of velocity is always in the direction of the motion. The changes his direction of motion with time, hence direction of velocity changes.

Conclusion:

Case (A)

At point A boy moves on a circular track with some constant velocity. The boy moves toward north direction.

Therefore the direction of boy at point A is North.

Case (B)

At point B, boy moves on a straight track having velocity. The boy moves towards west direction.

Therefore the direction of boy at point B is West.

Case (C)

At point C, the motion of boy is on the circular path. On the circular path the direction of velocity is in direction of motion. The boy moves in south direction.

Therefore the direction of boy is South.

Thus, the direction of velocity of boy at points A,B and C is $\text{North}$, $\text{West}$ and $\text{South}$ respectively.

(c)

To determine

The direction of velocity of boy at points A,B and C.

Answer to Problem 1OQ

The direction of velocity of boy at points A,B and C is $\text{West}$, $\text{Non-Existent}$ and $\text{East}$.

Explanation of Solution

The boy on whole track moves on two types of path, circular and straight path.

The moment the boy moves on circular path , the acceleration becomes perpendicular to the motion of the boy and it is directed towards its center.

The moment boy reaches the straight path the direction of acceleration is in the direction of velocity which is in the direction of motion.

Conclusion:

Case (A)

The boy moves in the circular path at point A. The direction of the velocity is in the north direction.

In circular motion, acceleration on the particle acts perpendicular to the direction of motion.

Therefore the direction of acceleration of the boy whose velocity is in north direction at point A will be in West direction

Case (B)

At point B, the boy moves in straight path. The boy moves with constant velocity. So the acceleration is not present and is zero from equation (III).

Therefore the direction of the acceleration at point b will not be present. The direction of acceleration is nonexistent.

Case (C)

At point C, the boy is in circular path. The direction of velocity of boy is in south direction.

In circular motion, direction of velocity is perpendicular to the direction of velocity.

Therefore, the direction of acceleration of boy having velocity in south direction at point C will have direction of acceleration in East direction.

Thus, the direction of velocity of boy at points A, B and C is $\text{West}$, $\text{Non-Existent}$ and $\text{East}$.