### Tangential and Normal Component of Acceleration

• Calculate required parameters - such as curvature and radius of curvature - for motion along plane curved path.

• Find tangential and normal components of acceleration

### Solved Example:

#### 28-2-01

A body is moving along a circular path with variable speed. It has:

Solution:

Since the body has variable acceleration, it has tangential acceleration. Also, since the body is moving on a circular path, that means it is continuously changing its direction, so it has centripetal acceleration acting towards the center of curvature. That means it has both tangential and radial accelerations.

### Solved Example:

#### 28-2-02

A particle is under uniform circular motion. Which is the wrong statement regarding its motion?

Solution:

Since the particle is under UNIFORM circular motion, it will have no tangential acceleration, only radial (normal) acceleration will be present. Its velocity will be towards the tangent. Hence velocity and acceleration will be perpendicular to each other. (valid only for UNIFORM circular motion.) Let us see each statement individually.

• The velocity vector is tangential to the circle: CORRECT

• The acceleration vector is tangential to the circle. WRONG since the motion is uniform, there is NO tangential acceleration, but it will have a centripetal acceleration which is towards the center of the curvature. So the acceleration vector is NORMAL or RADIAL.

• The acceleration vector is directed towards the centre of the circle. CORRECT

• The velocity and acceleration vectors are perpendicular to each other. CORRECT

### Solved Example:

#### 28-2-03

A particle revolves round a circular path with uniform angular velocity. The acceleration of the particle is:

Solution:

The key word is UNIFORM, hence tangential acceleration is absent, only radial acceleration will be there.

### Solved Example:

#### 28-2-04

A car is moving with speed 30 m/s on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/s$^2$. The acceleration of the car is:

Solution:
$a_{t}=2m/s^{2}$ $a_{n}=\dfrac {v^{2}}{r}=\dfrac {30^{2}}{500}=1.8m/s^{2}$ $a=\sqrt {a^{2}_{n}+a^{2}_{t}}=\sqrt {2^{2}+1.8^{2}}=2.69\ m/s^{2}$

### Solved Example:

#### 28-2-05

If cycle wheel of radius 0.4m completes 1 rev in 1 sec, then acceleration of the cycle is:

Solution:
$a=\dfrac {v^{2}}{r}=r\omega ^{2}$ $\omega =\dfrac {1\ rev}{1\ sec}=2\pi\ rad/s$ $a=\left( 0.4\right) \left( 2\pi \right) ^{2}=1.6\pi ^{2}m/s^{2}$

velocity at any height h from the top is given by: $v = \sqrt{2gh}$
$\mathrm{centrifugal\ force} = \dfrac{mv^2}{R} = \dfrac{2mgh}{R}$ Refer to the free body diagram as shown, $\dfrac{mv^2}{R} + N = mg \sin \theta$ The particle will leave the circle when normal reaction N = 0