Name: Conservation of Linear Momentum
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: In their study of natural phenomena over many centuries, scientists have identified certain fundamental laws. One such law is the law of conservation of mome...

Question

A car is speeding up on a horizontal road with an acceleration $a .$ Consider the following situations in the car.

(a) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle.

(b) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.

Accepted Answer

(a) Above diagram shows the ball suspended from the ceiling of an accelerated car with an acceleration $a .$ Being accelerated, the car is a non-inertial frame and it is not valid to apply Newton's law of motion in it, unless we apply a pseudo force on the object of observation in the opposite direction of the acceleration of motion.

In the figure shown above, let the inclination of the string with the vertical be $θ$ and the mass of the ball be $m$ then, $m a$ is the pseudo force. Other forces on the ball are its weight and the tension in the string.

Since, the ball is said to be in rest, on balancing the forces on it in two mutually perpendicular directions, we get,

$T \cos θ = m g$ and $T \sin θ = m a$

Dividing these two equations, we get,

$\tan θ = \frac{a}{g} \Rightarrow θ = \tan^{- 1} (\frac{a}{g})$

(b) Above diagram shows the block kept on a smooth incline in a car moving with an acceleration $a .$ Being accelerated, the car is a non-inertial frame and it is not valid to apply Newton's law of motion in it, unless we apply a pseudo force on the object of observation in the opposite direction of the acceleration of motion.

Let the inclination of the incline be $θ .$ Also, let the mass of the block be $m$ then, $m a$ will be the pseudo force. Other forces on the block are its weight and the normal due to the inclined surface.

Since, the block is said to be in rest, on balancing the forces on it in two mutually perpendicular directions, we get,

$N \cos θ = m g$ and $N \sin θ = m a$

Dividing these two equations, we get,

$\tan θ = \frac{a}{g} \Rightarrow θ = \tan^{- 1} (\frac{a}{g})$

Important Questions on Newton's Laws of Motion

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