A coin is placed 33 cm from the center of a
horizontal turntable, initially at rest. The
turntable then begins to rotate. When the
speed of the coin is 100 cm/s (rotating at a
constant rate), the coin just begins to slip.
The acceleration of gravity is 980 cm/s2 .
What is the coefficient of static friction be-
tween the coin and the turntable?
To find the coefficient of static friction between the coin and the turntable, we can use the equation for centripetal force.
The centripetal force acting on the coin is given by the equation:
F = m * a
where F is the centripetal force, m is the mass of the coin, and a is the centripetal acceleration.
The centripetal acceleration can be calculated using the equation:
a = (v^2) / r
where v is the velocity of the coin and r is the radius of the turntable (i.e., the distance from the center of the turntable to the coin).
In this case, the centripetal force is provided by the frictional force between the coin and the turntable, which is given by:
F = (coefficient of friction) * (normal force)
The normal force is equal to the weight of the coin, which can be calculated as:
m * g
where m is the mass of the coin and g is the acceleration due to gravity.
Setting these equations equal to each other, we have:
(coefficient of friction) * (m * g) = m * (v^2) / r
Simplifying, we get:
(coefficient of friction) = (v^2) / (g * r)
Plugging in the values given in the problem, we get:
(coefficient of friction) = (100^2) / (980 * 33)
Calculating this, we find:
(coefficient of friction) ≈ 0.101
Therefore, the coefficient of static friction between the coin and the turntable is approximately 0.101.