1. In first case we have the box with weight, say, 10 Newtons on incline. We calculate two components (perpendicular and parallel)of weight. We calculate normal force and force of friction. The box is at rest.
2. In second case we have ball on incline, made of same material as box from first case. Weight of the ball is the same as in first case, components of weight are the same as in first case. Normal force and friction have the same amount as in the case of box. But,unlike the box, ball is moving – it slides down. How’s that possible? If calculated net force is zero?
It is not true that the ball slides, but it will roll down.
In the first case, the forces balance each other and there is no motion.
In the second force, the frictional force acting tangentially backwards (up the incline) and the component of weight of the ball acting parallel to the plane and through the centre of the ball constitute a couple and tends to rotate it. Now there is no sliding; it rolls.
(If any further clarification is required please post as comment to this post)
Please refer to http://www.real-world-physics-problems.com/rolling-without-slipping.html for detailed treatment of the Physics of rolling without friction.
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an unloaded bus can be stopped by applying brakes on straight road after covering a distance x. Suppose, the passenger add 50% of its weight as the load and the braking force remains uncharged, how far will the bus go after the application of brakes? (velocity of bus in both case is same)
Say True or False; An object on the Moon weights 1/6 of what it weights on the Earth. Therefore, the object has less inertia on the Moon.( Justify your answer)
Answer: The statement is false.
Inertia depends on mass and not on weight. The weight on moon is less not because of a decrease in mass, but due to the decrease in the acceleration due to gravity on the surface of moon.
Inertia depends on mass and there is no change in mass when a body is taken to moon. Therefore the inertia of a body is same on moon as that on earth or anywhere else in the universe, even in a gravity free space.
Derive an expression for the minimum horizontal velocity to be given to a ball hanging vertically from a point so that it is able to just complete a vertical circular path.
While pushing, we are giving the ball some kinetic energy which will be converted to potential energy as it moves upward. If the whole kinetic energy is converted to potential energy at the top most point, it will fall straight down, resulting in only a semicircle. For the ball to continue the path, it should have a centrifugal force equal to weight of the body when it is at the topmost point.
let V be the initially applied velocity and v be the velocity at the topmost point
mv²/r = mg
v²/r = g
v² = rg
v = √(rg)
Potential Energy at the topmost point , Ep = 2mgr
Ep is the difference in kinetic energy between initial point(Ei) and at topmost point(Et).
Ei – Et = Ep
½mV² - ½mv² = 2mgr
½m(V²-v²) = 2mgr
V²-v² = 4gr
V² = 4gr + v²
V² = 4gr + rg
V² = 5rg