Unit 2: Force and Translational Dynamics
Weightage : 20-15%
The Gravitron is a carnival ride that looks like a large cylinder. People stand inside the cylinder against the wall as it begins to spin. Eventually, it rotates fast enough that the floor can be removed without anyone falling.
Given that the coefficient of friction between a person’s clothing and the wall is \(\mu\), the tangential speed is \(v\), and the radius of the ride is \(r\), what is the greatest mass that a person can be to safely go on this ride?
Correct Answer: \( \boxed{\mathrm{E}} \)
The upward static friction prevents the person from sliding down, while the wall’s normal force provides the required centripetal force.
At the threshold of slipping,
\(f_{\mathrm{s}}=mg\)
Since
\(f_{\mathrm{s}}=\mu N\)
and
\(N=\dfrac{mv^2}{r}\),
we obtain
\(mg=\mu\left(\dfrac{mv^2}{r}\right)\)
Cancelling the mass gives
\(g=\dfrac{\mu v^2}{r}\)
The mass cancels completely from the equation. Therefore, whether a person remains against the wall depends only on the coefficient of friction, the speed of the ride, and its radius.
Since no maximum mass can be determined from the given quantities, none of the expressions in choices (A) through (D) is correct.
Therefore, the correct answer is (E).
In a carnival ride, people of mass \(m\) are whirled in a horizontal circle by a floorless cylindrical room of radius \(r\), as shown in the diagram above. If the coefficient of friction between the people and the tube surface is \(\mu\), what minimum speed is necessary to keep the people from sliding down the walls?
Correct Answer: \( \boxed{\mathrm{B}} \)
The person remains at rest vertically, so the upward static friction balances the person’s weight.
\(f_{\mathrm{s}}=mg\)
The maximum static friction is
\(f_{\mathrm{s,max}}=\mu N\)
The normal force is provided by the wall and acts as the centripetal force:
\(N=\dfrac{mv^2}{r}\)
At the minimum speed required to prevent slipping,
\(mg=\mu\left(\dfrac{mv^2}{r}\right)\)
Solving for the speed,
\(v=\sqrt{\dfrac{rg}{\mu}}\)
Thus, the rider will not slide down only if the ride spins at or above this speed.
Therefore, the correct answer is (B).
A block of mass \(m\) is accelerated across a rough surface by a force of magnitude \(F\) that is exerted at an angle \(\phi\) with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude \(f\).
What is the acceleration of the block?
(A) \(\frac{F}{m}\)
(B) \(\frac{F\cos\phi}{m}\)
(C) \(\frac{F-f}{m}\)
(D) \(\frac{F\cos\phi-f}{m}\)
(E) \(\frac{F\sin\phi-mg}{m}\)
Correct Answer: \(\boxed{\mathrm{D}}\)
Resolve the applied force into its horizontal and vertical components.
The horizontal component of the applied force is
\(F_x=F\cos\phi\)
Friction acts opposite the motion with magnitude \(f\). Therefore, the net horizontal force is
\(\sum F_x=F\cos\phi-f\)
Applying Newton’s Second Law,
\(F\cos\phi-f=ma\)
Hence,
\(a=\frac{F\cos\phi-f}{m}\)
