Problem 4: A block of mass m slides down a frictionless incline as:
The block is released from height h above the bottom of the circular loop.
(a) Find the force exerted on the block by the inclined track at point A. (Hint: Consider the Newton’s 2nd law in the radial direction and you may use polar coordinates for this circular part of the path to relate the acceleration and velocity components).
(b) Find the force exerted on the block by the inclined track at point B.
(c) Find the speed of the block at point B.
(d) What is the maximum height that the block can reach after leaving the track?
(e) Find the distance between point A and the point that the block land on ground level?
To solve this problem, we can use concepts of circular motion and conservation of energy. Let's go through each part of the problem step by step:
(a) To find the force exerted on the block by the inclined track at point A, we need to consider the forces acting on the block at that point. The only force acting in the radial direction is the normal force from the track. At this point, the normal force provides the centripetal force required for circular motion. By using Newton's second law in the radial direction, we can equate the centripetal force to the mass times the acceleration. In polar coordinates, the radial acceleration (aᵣ) can be related to the velocity (v) and the angular velocity (ω) as aᵣ = v²/r, where r is the radius of the circular path. Therefore, the force exerted by the track can be found using the equation F = m * aᵣ = m * v²/r.
(b) At point B, the block is no longer in contact with the inclined track. Thus, there is no force exerted on the block by the track at this point. However, there will still be gravitational force acting on the block downward.
(c) To find the speed of the block at point B, we can use the principle of conservation of energy. At point A, we have gravitational potential energy due to the block's height above the bottom of the circular loop. As the block slides down, this potential energy is converted into kinetic energy. At point B, all the initial potential energy is converted into kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy to find the speed at point B. The equation is m * g * h = 1/2 * m * v², where g is the acceleration due to gravity.
(d) The maximum height that the block can reach after leaving the track can also be found using the conservation of energy principle. At the highest point, all the initial kinetic energy is converted into potential energy. Therefore, we can equate the initial kinetic energy to the final potential energy to find the maximum height.
(e) To find the distance between point A and the point where the block lands on the ground, we need to consider the horizontal motion of the block after leaving the track. Since there is no horizontal force acting on the block, it will continue moving with a constant horizontal velocity. We can find the time it takes for the block to reach the ground using the vertical motion equations, and then multiply this time by the horizontal velocity to find the distance.
By following these steps and utilizing the necessary equations, you should be able to find the answers to all parts of the problem.
(a) To find the force exerted on the block by the inclined track at point A, we can consider the forces acting on the block at that point. The only force acting on the block at point A is the normal force provided by the inclined track.
Since the block is sliding down a frictionless incline, there is no frictional force acting on it. Therefore, the force exerted on the block by the inclined track at point A is equal to the normal force.
(b) To find the force exerted on the block by the inclined track at point B, we again consider the forces acting on the block at that point. At point B, the block is inside a circular loop.
The forces acting on the block at point B are the normal force provided by the inclined track and the downward force due to gravity. The force exerted on the block by the inclined track is in the upward direction, which balances the downward force due to gravity. Therefore, the force exerted on the block by the inclined track at point B is equal to the normal force.
(c) To find the speed of the block at point B, we can use the conservation of mechanical energy. Since the incline is frictionless, mechanical energy is conserved.
Initially, at point A, the block has gravitational potential energy, given by mgh, since it is at a height h above the bottom of the circular loop. At point B, all of this potential energy is converted into kinetic energy, given by (1/2)mv^2, where v is the velocity of the block at point B.
Setting the gravitational potential energy equal to the kinetic energy, we have:
mgh = (1/2)mv^2
Simplifying the equation, we get:
gh = (1/2)v^2
Solving for v, we find:
v = sqrt(2gh)
(d) To find the maximum height that the block can reach after leaving the track, we need to consider the conservation of mechanical energy.
At the bottom of the circular loop, all of the block's mechanical energy is in the form of kinetic energy. When the block reaches its maximum height, all of this kinetic energy is converted into gravitational potential energy.
Using the conservation of mechanical energy:
(1/2)mv^2 = mgh
Simplifying the equation, we find:
v^2 = 2gh
Solving for h, we get:
h = (v^2)/(2g)
(e) To find the distance between point A and the point that the block lands on the ground level, we need to consider the projectile motion of the block after it leaves the track.
Since the only force acting on the block after it leaves the track is the downward force due to gravity, its motion can be treated as projectile motion.
Using the equations of projectile motion, we can find the horizontal distance traveled by the block, which is the distance between point A and the point that the block lands on the ground level.
The horizontal distance traveled by the block can be given by the equation:
d = v * t
where d is the horizontal distance, v is the horizontal component of velocity of the block, and t is the time of flight.
Since there is no horizontal force acting on the block, the horizontal component of velocity remains constant throughout the motion. Therefore, we can use the horizontal component of velocity at point B, which we already found in part (c).
t = (2v * sin(theta)) / g
where theta is the angle between the inclined track and the horizontal.
Substituting the value of t in the equation for d, we find:
d = v * ((2v * sin(theta)) / g)
Simplifying the equation, we get:
d = (2v^2 * sin(theta)) / g
Remember to substitute the values of v and theta in the equation to find the actual distance.