1) A block of mass 0.5 kg is placed at the highest point of a fixed smooth sphere
of radius 0.5m. The block is gently pushed to one side. Calculate: 0 the
3
o (1)
force exerted by the block on the sphere when it makes an angle
with the
vertical. (ii) At what angle does it leave the sphere ? (iii) At what distance
from the point of contact of the sphere does the block hit the ground?
To answer these questions, we need to consider the forces acting on the block and use the principles of circular motion.
(i) The force exerted by the block on the sphere when it makes an angle θ with the vertical can be calculated using Newton's second law. At any point on the sphere, there are two forces acting on the block: the weight (mg) acting vertically downward and the normal force (N) acting perpendicular to the surface of the sphere. The normal force provides the centripetal force needed to keep the block moving in a circular path.
The centripetal force is given by:
Fc = m * a
where m is the mass of the block and a is the centripetal acceleration. In this case, the centripetal acceleration is given by the angular acceleration (α) times the distance from the center of the sphere (r):
a = α * r
The angular acceleration (α) can be found using the relation between angular acceleration and tangential acceleration:
α = a_t / r
where a_t is the tangential acceleration, which can be calculated using the following equation:
a_t = r * ω^2
where ω is the angular velocity, which can be expressed as:
ω = dθ / dt
The force exerted by the block on the sphere is equal to the normal force (N) acting in the opposite direction. So, we have:
N = m * α * r
Substituting the value of α from the earlier equation, we get:
N = m * (r * ω^2) / r
Simplifying, we get:
N = m * ω^2
(ii) The block leaves the sphere when the normal force becomes zero. At that point, only the force of gravity is acting on the block. So, we can equate the weight to the centripetal force:
mg = m * ω^2
Canceling the mass, we get:
g = ω^2
Taking the square root of both sides, we have:
ω = sqrt(g)
The angle θ at which the block leaves the sphere is given by:
θ = 90° - ω * t
where t is the time taken to reach that point.
(iii) To find the distance from the point of contact where the block hits the ground, we need to calculate the time taken for the block to reach the ground.
The time taken for one full rotation (T) can be calculated using the formula:
T = 2π / ω
The time taken for the block to reach the ground can be expressed as a fraction of T:
t = k * T
where k is a constant between 0 and 1. The actual value of k can be determined by considering the initial conditions of the problem.
Finally, the distance from the point of contact where the block hits the ground can be calculated using the equation for displacement in uniform circular motion:
s = r * (1 - cos(θ))
where r is the radius of the sphere.
By plugging in the appropriate values and following the steps outlined above, you can calculate the force exerted by the block on the sphere, the angle at which it leaves the sphere, and the distance from the point of contact where it hits the ground.