Question
Question: A 62 kg woman is a passenger in a “rotor ride” at an amusement park. A d of \(5 m\) is spun with an ...
A 62 kg woman is a passenger in a “rotor ride” at an amusement park. A d of 5m is spun with an angular velocity of 25rpm. The woman is pressed against the wall of the rotating drum. While the drum rotates, the floor is lowered. A vertical static friction force supports the woman’s weight. What must the coefficient of friction be to support her weight?
(A) 0.412
(B) 0.292
(C) 0.135
(D) 0.5
Solution
The frictional force balances the weight of the body and frictional force is the product of coefficient of static friction and force due to reaction of the drum when the drum is rotating.
Complete step by step answer:
A woman of mass m = 62kg is a passenger in a rotor ride. A d =5m is spun with an angular velocity ω = 25rpm which means 25 revolutions in one minute. Hence, number of revolutions in one second = 6025
The angle covered in one revolution is 2π and the velocity is taken as angular velocity due to the rotation of the drum.
Thus, the angular velocity = angular displacement in 25 revolutions per unit time
angular velocity =2π×6025
angular velocity = 6050πrad/s
Now, the force (F)=mω2r
F=62×(6050π)2×5
⇒F=176400375100000
∴F=2126.41N
When the drum rotates, the floor is lowered and a vertical static friction supports the weight of the woman i.e., frictional force = weight of the woman
⇒μF=mg[ μ is the coefficient of static friction and g is acceleration due to gravity]
⇒μ×2126.41=62×10
⇒μ=2126.41620
⇒μ=0.2915
∴μ=0.292(rounded off)
Therefore, option B is correct.
Note: The frictional force is the product of normal reaction force and coefficient of static friction. The normal reaction force is provided by the drum i.e., mω2r. Also remember that frictional force is always opposite to the direction of motion of a body.