Skidding Of Object On A Rotating Platform

Skidding of Object on a Rotating Platform

Skidding of an object on a rotating platform occurs when the object loses its grip and slides outward due to the centrifugal force acting on it. As the platform rotates, the object experiences an outward force proportional to the square of its velocity and the radius of its position on the platform. If the frictional force between the object and the platform is insufficient to counteract this outward force, the object will skid.

$\begin{aligned} & \text { Centrifugal force } \leq \text { Force of friction } \\ & m \omega^2 r \leq \mu m g \\ & \therefore \omega_{\max }=\sqrt{\frac{\mu g}{r}}=\text { It is the maximum angular velocity of rotation of the platform so that the object will not skid on it. } \\ & \omega=\text { Angular velocity } \\ & \mathrm{r}=\text { radius } \\ & \mu=\text { coefficient of friction }\end{aligned}$

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Example 1: A block of mass m is placed on a rotating platform at a distance r from the axis of rotation. What should be the maximum angular velocity to avoid skidding of the block [ take $\mu$ =the coefficient between the block and rotating plateform]

1) $\sqrt{\mu r g}$
2) $\sqrt{\frac{\mu r}{g}}$
3) $\sqrt{\frac{\mu g}{r}}$
4) $\mu \sqrt{\frac{r}{g}}$

Solution:

To avoid skidding

centripetal force = force of friction

$\begin{aligned} & m w_{\max }^2 r=\mu m g \\ & w_{\max }=\sqrt{\frac{\mu g}{r}}\end{aligned}$

Hence, the answer is the option (3).

Example 2: A block of mass m is kept on the edge of the horizontal turn table of radius R. The Turn table is rotating with constant angular velocity $\omega$. coefficient of friction is $\mu$. If the block is just about to move find the angular velocity w of the turn table

$\begin{aligned} & \text { 1) } \sqrt{\frac{\mu g}{R}} \\ & \text { 2) } \sqrt{\frac{\mu}{R g}} \\ & \text { 3) } \sqrt{\frac{\mu}{R}} \\ & \text { 4) } \sqrt{\frac{R}{\mu g}}\end{aligned}$

Solution:

Skidding of the object on a Rotating Platform

Centripetal force $\leq$ Force of friction
$
\begin{aligned}
& m \omega^2 r \leq \mu m g \\
& \therefore \omega_{\max }=\sqrt{(\mu g / r)} \\
& \omega=\text { Angular velocity } \\
& \mathrm{r}=\text { radius } \\
& \mu=\text { coefficient of friction } \\
& \text { wherein }
\end{aligned}
$
It is the maximum velocity of rotation of the platform so that the object will not skid on it.
Centrifugal force on the block

$
F_c=m w^2 R
$

$
f_s=m w^2 R
$

for limiting case

$
f_s=f_L=\mu N=\mu m g
$

centripetal force will be provided by limiting frictional force

$
\mathrm{So}^{\mu m g}=m w^2 R \Rightarrow w=\sqrt{\frac{\mu g}{R}}
$

Hence, the answer is the option (1).

Summary

Skidding on a rotating platform might happen since the frictional force between that object and the platform may not be sufficient to produce the needed centripetal force to keep it in its curved path. Because the platform rotates, the centripetal force pulls the object toward the centre. However, should the friction become too low because of a very smooth surface, large speed, or a heavy mass of the object, then the inertia of the object will just slide it outward. This very effect is what makes things slide off a lazy Susan and what keeps people safe on amusement park rides where they would skid if not for the safety features built into their ride. Something must be done in order to ensure adequate friction to keep things from skidding. Non-slip surfaces or proper tethering of objects will do the trick. Knowing these concepts a rotating platform or ride can be designed and run much safer from accident.