Physics · Laws Of Motion · NEET
Friction holds the rider up, not the normal force. The wall pushes horizontally inward on the rider (this is the normal force N). Because the rider presses hard against the wall, a large upward friction force f = mu N acts along the wall. This friction balances the weight mg. So the rider is held up by friction, while gravity still pulls down. Gravity is not cancelled by the normal force here, because N is horizontal (inward) and mg is vertical (downward).
The horizontal normal reaction N from the wall provides the centripetal force. For circular motion the body needs an inward (center-pointing) force equal to m v^2 / R, or m omega^2 R. The only inward force is the wall pushing on the body, so N = m v^2 / R = m omega^2 R. Do not treat friction as centripetal here; friction is vertical (up), it only balances the weight.
Faster spin means a larger inward normal force N, which gives a larger maximum friction mu N. To hold the weight you need mu N >= mg. If the speed is too small, N is small, friction is small, and the rider slides down. So there is a MINIMUM speed v = sqrt(gR / mu). Any speed above this is safe, so there is no maximum. This is opposite to a car on a curve, where too much speed causes slipping.
They are different setups. In the Death Well the axis is vertical and the rider moves in a HORIZONTAL circle around the inside of the wall, so gravity is held by friction and N points sideways. In a vertical circle (loop or bucket) the plane is vertical, gravity acts along the radius at the top and bottom, and the condition uses v_top = sqrt(gR), not friction. Do not mix the two formulas.
No. Set mu N = mg with N = m v^2 / R. This gives mu (m v^2 / R) = m g, and mass m cancels on both sides. So v_min = sqrt(gR / mu) depends only on gravity g, radius R, and friction coefficient mu. A heavy and a light rider need the same minimum speed.
A block of mass 10 kg is in contact with the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the wall is 0.1. The minimum angular velocity of the drum (rotating about its vertical axis) needed to keep the block stationary is (g = 10 m/s^2):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The minimum linear speed is v_min = sqrt(gR / mu), where g is gravity, R is the radius of the well, and mu is the coefficient of friction between the rider and the wall. In terms of angular velocity it is omega_min = sqrt(g / (mu R)).
They are the same physics. The Death Well (or Wall of Death) is a fairground stunt where a rider circles the inside of a vertical wooden well. The Rotor is an amusement ride: a spinning drum where the floor drops away and riders stick to the wall. In all of them, friction from the wall balances gravity while the normal force keeps the rider in the circle.
If the angular velocity is below omega_min = sqrt(g / (mu R)), the inward normal force N becomes small, so the maximum friction mu N is smaller than the weight mg. Friction cannot hold the body, and it slides down the wall. That is why the ride must reach a minimum spin before the floor is dropped.
No. On a normal horizontal floor N = mg. But on a vertical wall, N is horizontal and equals m v^2 / R (the centripetal force). The weight mg is balanced by the vertical friction f = mu N, not by N itself. This is a common NEET mistake.
