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A horizontal circular turntable rotates about its center at the uniform rate of 10.9 revolutions per minutes. Find the greatest distance (in meter) from the center at which a small block will remain stationary relative to the turn table.The coefficient of static friction between the turntable and the object is 0.8.

Respuesta :

Answer:

6.03 m

Explanation:

First of all, let's convert the angular velocity from revolutions per minute to radians per second:

[tex]\omega = 10.9 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=1.14 rad/s[/tex]

The frictional force on the block ranges from zero to a maximum value of

[tex]F=\mu mg[/tex]

In order for the block to remain stuck on the turntable, the frictional force must be equal to the centripetal force, so we can write:

[tex]m\omega^2 r = \mu mg[/tex]

where

m is the mass of the block

[tex]\omega[/tex] is the angular velocity

r is the distance of the block from the centre

[tex]\mu = 0.8[/tex] is the coefficient of static friction

g = 9.8 m/s^2

Solving for r, we find:

[tex]r=\frac{\mu g}{\omega^2}=\frac{(0.8)(9.8)}{(1.14)^2}=6.03 m[/tex]

Horizontal circular turntable rotates about center at greatest distance from center at which a small block remain stationary relative to turn table is 6.03 meters.

What is centripetal force?

Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle towards the center of rotational path.

It can be given as,

F=mω²r

Here, (m) is the mass of the body and (r) is the radius of circular path.

Frictional force can be given as,

F=μmg

Here, (μ) is the coefficient of static friction.

Given information-

The rotation of circular turntable is 10.9 revolutions per minutes.

The coefficient of static friction between the turntable and the object is 0.8.

As 1 rpm equal to 0.10472 radians per second. Thus the angular velocity 10.9 revolutions per minutes is equal to 1.14 radians per second.

ω=1.14 rad/sec.

As the small block remain stationary relative to the turn table. Thus the frictional force must be equal to the centripetal force. Therefore,

mω²r=μmg

ω²r=μg

Put the values in the above equation (g=9.8 m/s²)we get the value of r as,

r=6.03 meters.

Thus, the greatest distance (in meter) from the center at which a small block will remain stationary relative to the turn table is 6.03 meters.

Learn more about the centripetal force here;

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