Advertisements
Advertisements
Question
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?
Advertisements
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 (Fg = mg).
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
RELATED QUESTIONS
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for `omega <= sqrt(g/R)` .What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega = sqrt("2g"/R)` ?Neglect friction.
When a particle moves in a circle with a uniform speed
A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car NA and NB when it is at the points A and B respectively.
A coin placed on a rotating turntable just slips. If it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, the tension is the string is mg cos θ
The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its
(a) velocity remains constant
(b) speed remains constant
(c) acceleration remains constant
(d) tangential acceleration remains constant.
A car of mass M is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing at a rate a.
(a) The acceleration of the car is towards the centre of the path.
(b) The magnitude of the frictional force on the car is greater than \[\frac{\text{mv}^2}{\text{r}}\]
(c) The friction coefficient between the ground and the car is not less than a/g.
(d) The friction coefficient between the ground and the car is \[\mu = \tan^{- 1} \frac{\text{v}^2}{\text{rg}.}\]
A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is cm/s and t in seconds.
(a) Find the radial acceleration of the particle at t = 1 s.
(b) Find the tangential acceleration at t = 1 s.
(c) Find the magnitude of the acceleration at t = 1 s.
The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this instant.
Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos θ ≈ 1 − θ2/2 and SINθ ≈ θ for small θ.
A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R (In the following figure). A smooth groove AB of length L(<<R) is made the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.
Choose the correct option.
Consider the following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit.
Principle of conservation of angular momentum comes in force in which of these?
A particle is moving in a radius R with constant speed v. The magnitude of average acceleration after half revolution is ____________.
Two identical masses are connected to a horizontal thin (massless) rod as shown in the figure. When their distance from the pivot is D, a torque τ produces an angular acceleration of α1. The masses are now repositioned so that they are 2D from the pivot. The same torque produces an angular acceleration α2 which is given by ______
A child starts running from rest along a circular track of radius r with constant tangential acceleration a. After time the feels that slipping of shoes on the ground has started. The coefficient of friction between shoes and the ground is [g = acceleration due to gravity].
If a cyclist doubles his speed while negotiating a curve, how does the tendency to overturn vary?
A body of mass m is performing a UCM in a circle of radius r with speed v. The work done by the centripetal force in moving it through `(2/3)`rd of the circular path is ______.
A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is ______.
A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is `sqrt(x("u"^2 - "gL")`. The value of x is ______.
Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh-1 and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s2)
Statement II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2 m radius with the speed of 18.5 kmh-1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
