A hollow sphere of radius 0.30 m, with rotational inertia I = 0.080 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 19° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 70 J.
(a) How much of this initial kinetic energy is rotational?
J
(b) What is the speed of the center of mass of the sphere at the initial position?
m/s
Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?
J
(d) What is the speed of its center of mass now?
m/s
4kg
To solve this problem, we need to use the principles of rotational motion and conservation of energy.
(a) To find the amount of initial kinetic energy that is rotational, we first need to find the total kinetic energy of the sphere. We can do this by using the formula:
Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy
Given that the total kinetic energy is 70 J and we are looking for the rotational part, we can write:
70 J = Translational Kinetic Energy + Rotational Kinetic Energy
Since the sphere is rolling without slipping, we know that the translational and rotational velocities are related. The translational kinetic energy can be expressed as (1/2) * mass * (velocity of center of mass)^2, and the rotational kinetic energy can be expressed as (1/2) * moment of inertia * (angular velocity)^2.
Since the rotational inertia (moment of inertia) I and the angular velocity ω are given, we can calculate the translational kinetic energy as follows:
Translational Kinetic Energy = Total Kinetic Energy - Rotational Kinetic Energy
Translational Kinetic Energy = 70 J - (1/2) * 0.080 kg m^2 * (angular velocity)^2
(b) To find the speed of the center of mass of the sphere at the initial position, we can use the relationship between translational velocity and angular velocity for a rolling object. The translational velocity v and angular velocity ω are related by the equation:
v = ω * radius
Given that the sphere rolls without slipping, we know that the point of contact on the surface is instantaneously at rest. Therefore, the angular velocity ω multiplied by the radius of the sphere will give us the speed of the center of mass.
v = ω * radius
v = ω * 0.30 m
Now, we need to find the angular velocity ω from the given rotational inertia I. The formula for rotational inertia of a solid sphere is:
I = (2/5) * mass * radius^2
In this case, we can solve for ω as follows:
I = 0.080 kg m^2
I = (2/5) * mass * (0.30 m)^2
0.080 kg m^2 = (2/5) * mass * (0.30 m)^2
Solve the above equation to find the mass.
Once you have the mass, you can substitute it back into the equation v = ω * 0.30 m to find the speed of the center of mass.
(c) To find the total kinetic energy after the sphere has moved 1.0 m up the incline from its initial position, we need to consider the work done against the gravitational force as the sphere moved up the incline. The potential energy gained is equal to the work done against gravity.
Potential Energy = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height the sphere has moved up the incline (which is given as 1.0 m).
The total kinetic energy can be calculated as the sum of the kinetic energy due to translation and rotation:
Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy
We can use the same formulas as in part (a) to calculate these two components and then determine the total kinetic energy.
(d) To find the speed of the center of mass now, we can use the relationship between translational velocity and angular velocity as explained in part (b). The translational velocity v and angular velocity ω are related by the equation:
v = ω * radius
Substitute the given values and solve for v.