# 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 - Physics

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 1

When the motorcyclist is at the highest point of the death-well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him

:. R + mg = (mv^2)/r ......(1)

Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass of the motorcycle). Because of the balancing of the forces, the motorcyclist does not fall down.

The minimum speed required to perform a vertical loop is given by equation (1) when R = 0.

:. mg = mv_"min"^2  or v_"min"^2 = gr

or v = sqrt(gr) = sqrt(10xx25) ms^(-1) = 15.8 ms^(-1)

#### Solution 2

In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (FN) and the force due to gravity (Fmg).

The equation of motion for the centripetal acceleration ac, can be written as:

Fnet = mac

F_N + F_g= ma_c

F_N + mg =- (mv^2)/r

Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (vmin), FN = 0

mg = mv_"min"^2

:.V_min = sqrt(rg)

= sqrt(25xx10) = 15.8 m/s

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 5 Laws of Motion
Q 38 | Page 113