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Q5.38:
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m ?

Solution :

Step 1: Theoretical Analysis of Forces at the Uppermost Point

At the highest point of a vertical loop of radius $R$, two primary forces act on the motorcyclist (of combined mass $m$):

  • The gravitational force ($F_g = mg$), acting vertically downward.
  • The normal reaction ($N$) exerted by the wall of the death-well, also acting vertically downward.

[Per Newton's Second Law of Motion], the net centripetal force ($F_c$) required to maintain circular motion is provided by the sum of these radial forces:

$F_c = \frac{mv^2}{R} = N + mg$

Step 2: Condition for Maintaining Contact

For the motorcyclist to avoid dropping, the normal reaction $N$ must be $\ge 0$. If $N$ becomes negative, the motorcyclist loses contact with the surface. The critical condition (minimum speed) occurs when the motorcyclist is on the verge of losing contact, i.e., $N = 0$.

Substituting $N = 0$ into the force equation:

$\frac{mv_{min}^2}{R} = 0 + mg$

Simplifying for $v_{min}$:

$v_{min}^2 = Rg$

$v_{min} = \sqrt{Rg}$

Step 3: Visual Representation of Forces

mg N Top Point

Step 4: Calculation with Given Parameters

Given: Radius $R = 25\text{ m}$, Acceleration due to gravity $g \approx 9.8\text{ m/s}^2$.

Substituting values into the derived formula:

$v_{min} = \sqrt{25 \times 9.8}$

$v_{min} = \sqrt{245}$

$v_{min} \approx 15.65\text{ m/s}$

Explanation: The motorcyclist does not fall because at the critical velocity, the downward acceleration required to keep the body in a circular path is exactly equal to the acceleration due to gravity ($g$). The weight of the rider provides the necessary centripetal force to maintain the loop, effectively making the rider "weightless" with respect to the surface at that precise speed.

Final Solution: The minimum speed required is $15.65\text{ m/s}$.


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