3 2 Centripetal Force – Class 11TH Physics

3.2 Centripetal Force

Newton's second law of motion states that when a force acts on a body, it produces acceleration in the same direction. A force acting on a moving body along the direction of its velocity will change magnitude of the velocity (speed) keeping the direction unchanged. On the other hand, a constant force acting perpendicular to the velocity of a body moving in a circular path will change the direction but magnitude of velocity (speed) will remain the same. Such force makes the body move in a circle by producing a radial (or centripetal) acceleration and is called centripetal force (Centre seeking) force. Figure 3.6(a) shows a ball tied at the end of a string is whirled in a horizontal surface. It would not continue in a circular path if the string is snapped. Careful observation shows at once that if the string snaps, when the ball is at the point $A$ , in Fig. 3.6 (b), the ball will follow the straight line path $AB$ which is tangent $AB$ at point $A$ .

Thus, a force is needed to change the direction of velocity or motion of a body continuously at each point in circular motion moving with uniform speed. The force that does not alter speed but only direction at each point is a perpendicular force which acts along the radius of the circular path. This force always pulls the object towards the centre of the circular path. Its direction is perpendicular to the tangential velocity at each point.

The force needed to bend the straight path of the particle into a circular path is called the centripetal force.

For a body of mass $m$ moving with velocity $v$ in a circular path of radius $r$ , centripetal force $F_{c}$ is given by:

F_{c} = ma_{c} = \frac{mv^{2}}{r} \quad \ldots (3.12)

where $a = v^{2}/r$ is the centripetal acceleration and its direction is towards the centre of the circle. As $v = r\omega$ , so the above equation becomes:

F_{c} = mr\omega^{2} \quad \ldots (3.13)

Example 3.2

If a CD spins at $210\,\mathrm{rpm}$ , what is the radial acceleration of a point on the outer rim of the CD? The CD is $12\,\mathrm{cm}$ in diameter.

Solution

We convert $210\,\mathrm{rpm}$ into a frequency in revolutions per second (Hz).

Thus

f = \frac{210\,\mathrm{rpm}}{60\,\mathrm{s}} = 3.5\,\mathrm{rev/s} = 3.5\,\mathrm{Hz}

For each revolution, the CD rotates through an angle of $2\pi$ radians. The angular velocity is:

\omega = 2\pi f = 2\pi \times 3.5\,\mathrm{s^{-1}} = 7.0\pi\,\mathrm{rad\,s^{-1}}

The radial acceleration is:

a = \omega^{2}r = (7.0\pi\,\mathrm{rad\,s^{-1}})^{2} \times 0.06\,\mathrm{m} = 29\,\mathrm{m\,s^{-2}}

Example 3.3

A ball tied to the end of a string, is swung in a vertical circle of radius $r$ under the action of gravity as shown in Fig. 3.7. What will be the tension in the string when the ball is at the point $A$ of the path and its speed is $v$ at this point?

Solution

For the ball to travel in a circle, the force acting on the ball must provide the required centripetal force. In this case, at point $A$ , two forces act on the ball, the pull of the string and the weight of the ball. These forces act along the radius at $A$ , and so their vector sum must furnish the required centripetal force. We, therefore, have:

T + w = \frac{mv^{2}}{r}

as $w = mg$ , therefore:

T = \frac{mv^{2}}{r} - mg = m\left(\frac{v^{2}}{r} - g\right)

If $v^{2}/r = g$ , then $T$ will be zero and the centripetal force is just equal to the weight.