A block of mass m is kept on a horizontal tur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For the block not to slip,the required centripetal force must be provided by the static friction force.The condition for the block not to slip is:

Vedclass · Accepted Answer

$\sqrt{\frac{\mu g}{x}}$

Vedclass · Answer

(B) For the block not to slip,the required centripetal force must be provided by the static friction force.The condition for the block not to slip is:$f_{s} \leq \mu N$Here,the centripetal force is $F_{c} = m \omega^2 x$.The normal force $N$ on the horizontal surface is $N = mg$.Therefore,the maximum static friction force is $f_{s,max} = \mu mg$.Equating the centripetal force to the maximum static friction force:$m \omega^2 x = \mu mg$Dividing both sides by $mx$:$\omega^2 = \frac{\mu g}{x}$Taking the square root of both sides,we get the maximum angular speed:$\omega = \sqrt{\frac{\mu g}{x}}$

$A$ block of mass $m$ is kept on a horizontal turntable at a distance $x$ from the center. If the coefficient of friction between the block and the surface of the turntable is $\mu$,what is the maximum angular speed of the table so that the block does not slip?

Similar Questions

On a dry road,the maximum speed of a vehicle along a circular path is $V$. When the road becomes wet,the maximum speed becomes $\frac{V}{2}$. If the coefficient of friction of the dry road is $\mu$,then the coefficient of friction of the wet road is:

If a cyclist moving with a speed of $4.9 \, m/s$ on a level road can take a sharp circular turn of radius $4 \, m$,then the coefficient of friction between the cycle tyres and the road is:

$A$ small object placed on a rotating horizontal turntable just slips when it is placed at a distance $4\, cm$ from the axis of rotation. If the angular velocity of the turntable is doubled,the object slips when its distance from the axis of rotation is (in $, cm$)

$A$ motorcyclist has to rotate in horizontal circles inside the cylindrical wall of inner radius '$R$' metre. If the coefficient of friction between the wall and the tyres is '$\mu_{s}$',then the minimum speed required is ($g=$ acceleration due to gravity).

The maximum velocity (in $m/s$) with which a car driver must traverse a flat curve of radius $150\, m$ and coefficient of friction $0.6$ to avoid skidding is ........ $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module