1.Asolid sphere has amoment of inertia of 0.8kgm2&diameter 10cm rolls down along an inclined plane of length 80cm which is inclined at 37` to the horizontal .What is the transitional momentum of the sphere as it reaches to the foot of the inclined plane?
To find the translational momentum of the sphere as it reaches the foot of the inclined plane, we need to consider the conservation of mechanical energy.
The mechanical energy of the sphere at the top of the inclined plane is equal to the sum of its potential energy (PE) and its kinetic energy (KE).
The potential energy (PE) can be calculated using the formula:
PE = m * g * h
Where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the inclined plane. Since the height is not given, we need to find it.
Given that the inclined plane is inclined at an angle of 37 degrees to the horizontal and has a length of 80 cm, we can use trigonometry to find the height.
The height (h) can be calculated using the formula:
h = L * sin(θ)
Where L is the length of the inclined plane and θ is the angle of inclination.
Substituting the values, we get:
h = 0.8 * sin(37°) = 0.4774 m
Now we can calculate the potential energy (PE):
PE = m * g * h = 0.8 * 9.8 * 0.4774 = 3.745 J
Since there is no external force acting on the sphere as it rolls down the inclined plane (neglecting friction), the mechanical energy is conserved, meaning the potential energy at the top is equal to the kinetic energy at the bottom.
So the total mechanical energy at the bottom is 3.745 J.
The kinetic energy (KE) can be calculated using the formula:
KE = (1/2) * I * ω^2
Where I is the moment of inertia of the sphere and ω is the angular velocity of the sphere.
The moment of inertia (I) of a solid sphere is given by the formula:
I = (2/5) * m * r^2
Where m is the mass of the sphere and r is the radius of the sphere.
Given that the diameter of the sphere is 10 cm, the radius (r) is 5 cm or 0.05 m.
Substituting the values, we get:
I = (2/5) * 0.8 * (0.05)^2 = 0.00032 kg·m^2
Since the sphere is rolling without slipping, the linear velocity (v) of the sphere can be calculated using the formula:
v = ω * r
Given that the sphere is rolling down the inclined plane without slipping, we can find the angular velocity (ω) using the relation between the linear velocity and angular velocity:
v = rω
Substituting the values, we get:
v = 0.05ω
Since the sphere is rolling down the inclined plane, its linear velocity (v) can be calculated using the equation of motion:
v^2 = u^2 + 2as
Where u is the initial velocity (which is 0, as the sphere starts from rest), a is the acceleration, and s is the distance traveled down the inclined plane.
Given that s is 80 cm or 0.8 m and the acceleration (a) can be calculated using the formula:
a = g * sin(θ)
Substituting the values, we get:
a = 9.8 * sin(37°) = 5.892 m/s^2
Now we can calculate the final velocity (v):
v^2 = 0 + 2 * 5.892 * 0.8 = 9.4272 m^2/s^2
Taking the square root, we get:
v = √9.4272 = 3.0703 m/s
Since v = ω * r, we can find the angular velocity (ω):
ω = v / r = 3.0703 / 0.05 = 61.406 rad/s
Now we can calculate the kinetic energy (KE):
KE = (1/2) * 0.00032 * (61.406)^2 = 0.3632 J
Since the total mechanical energy at the bottom is equal to the sum of the potential energy and kinetic energy, we can find the translational momentum by equating it to the kinetic energy:
3.745 = 0.3632 + Translational momentum
Simplifying, we get:
Translational momentum = 3.745 - 0.3632 = 3.3818 J
Therefore, the translational momentum of the sphere as it reaches the foot of the inclined plane is 3.3818 J.