A solid sphere of radius 3 cm and mass of 100 g rolls down a 30 degree incline from a height of 1 m. A hoop of the same mass and radius rolls down the same incline from the same height. What is the linear velocity of each body at the bottom of the incline? From what height would the body with the larger moment of inertia have to roll down from so that the two linear velocities at the bottom of the incline would be the same
To find the linear velocity of each body at the bottom of the incline, we can use the principles of rotational and translational motion.
1. Solid sphere:
Let's start with the solid sphere. We can calculate its linear velocity using the conservation of mechanical energy. At the top of the incline, the initial potential energy is equal to the sum of the final kinetic energy (rotational and translational) and any energy losses due to friction.
Step 1: Calculate potential energy at the top
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
PE = 0.1 kg * 9.8 m/s^2 * 1 m = 0.98 J
Step 2: Calculate the moment of inertia (I) of the solid sphere
The moment of inertia for a solid sphere is given by I = (2/5) * m * r^2, where r is the radius and m is the mass.
I = (2/5) * 0.1 kg * (0.03 m)^2 = 2.4 x 10^-4 kg * m^2
Step 3: Calculate the angular velocity (ω) of the solid sphere at the bottom of the incline using the conservation of mechanical energy:
PE = translational kinetic energy + rotational kinetic energy
0.98 J = (1/2) * m * v^2 + (1/2) * I * ω^2
Since the solid sphere is rolling, we know that v = ω * r, where v is the linear velocity and r is the radius of the sphere.
0.98 J = (1/2) * m * v^2 + (1/2) * (2.4 x 10^-4 kg * m^2) * (v^2 / r^2)
0.98 J = (1/2) * (0.1 kg) * v^2 + (1/2) * (2.4 x 10^-4 kg * m^2) * (v^2 / (0.03 m)^2)
Now you can solve for v using algebraic manipulation.
2. Hoop:
The calculation for the hoop will be similar to the solid sphere, but the moment of inertia for a hoop is different. The moment of inertia of a hoop is given by I = m * r^2.
Step 1: Calculate the moment of inertia (I) of the hoop
I = m * r^2 = (0.1 kg) * (0.03 m)^2 = 9 x 10^-5 kg * m^2
Step 2: Use the conservation of mechanical energy to find the angular velocity (ω) at the bottom of the incline:
0.98 J = (1/2) * m * v^2 + (1/2) * I * ω^2
0.98 J = (1/2) * (0.1 kg) * v^2 + (1/2) * (9 x 10^-5 kg * m^2) * (v^2 / (0.03 m)^2)
Again, solve for v using algebraic manipulation.
To find the height from which the body with the larger moment of inertia would need to roll down so that the two linear velocities at the bottom of the incline are the same, you would equate the linear velocity equations for the sphere and the hoop and solve for the height.