A solid disk (mass = 1 kg, R=0.5 m) is rolling across a table with a translational speed of 9 m/s.
a.) What must the angular speed of the disk be?
? rad/s
b.) What is the rotational KE of the disk?
? J
c.) What is the total KE of the disk?
? J
d.) The disk then rolls up a hill of height 2 m, where the ground again levels out. Find the translational and rotational speeds now.
? m/s
? rad/s
I will be happy to critique your thinking.
For some reason everytime I try to calculate the KE rotational I do not get the right answer. The translational speed can be used for V, right?
To answer these questions, we can use the equations related to rotational motion and kinetic energy.
a.) To find the angular speed of the disk, we can use the relationship between linear speed (V) and angular speed (ω) for a rolling object. The formula is given as:
V = R * ω
where V is the linear speed, R is the radius of the disk, and ω is the angular speed.
In this case, the linear speed is given as 9 m/s, and the radius of the disk is 0.5 m. By substituting these values into the equation, we can solve for the angular speed:
9 m/s = 0.5 m * ω
Dividing both sides of the equation by 0.5 m, we have:
ω = 18 rad/s
Therefore, the angular speed of the disk must be 18 rad/s.
b.) The rotational kinetic energy (KE) of the disk can be calculated using the formula:
Rotational KE = (1/2) * I * ω^2
where I is the moment of inertia of the disk and ω is the angular speed.
The moment of inertia for a solid disk rotating about its central axis is given by the equation:
I = (1/2) * m * R^2
where m is the mass of the disk and R is its radius.
Given that the mass of the disk is 1 kg and the radius is 0.5 m, we can substitute these values into the equation to find the moment of inertia:
I = (1/2) * 1 kg * (0.5 m)^2 = 0.125 kg * m^2
Now we can substitute the values of the moment of inertia and the angular speed into the formula for rotational KE:
Rotational KE = (1/2) * 0.125 kg * m^2 * (18 rad/s)^2 = 22.5 J
Thus, the rotational kinetic energy of the disk is 22.5 Joules.
c.) The total kinetic energy (KE) of the disk is the sum of its translational KE and rotational KE. The translational KE is given by the formula:
Translational KE = (1/2) * m * V^2
where V is the linear speed of the disk.
Given that the mass of the disk is 1 kg and the linear speed is 9 m/s, we can substitute these values into the formula to find the translational KE:
Translational KE = (1/2) * 1 kg * (9 m/s)^2 = 40.5 J
Now we can add the translational KE and rotational KE to get the total KE:
Total KE = 40.5 J + 22.5 J = 63 J
Therefore, the total kinetic energy of the disk is 63 Joules.
d.) When the disk rolls up a hill of height 2 m, its total energy is conserved. This means that the total kinetic energy at the start (before the hill) is equal to the total kinetic energy at the top of the hill (after the hill).
Since the translational and rotational KE contributed to the total KE before the hill, they also contribute to the total KE after the hill.
We can use this information to find the translational and rotational speeds after the hill.
Let's assume the translational speed after the hill is V2, and the angular speed is ω2.
Using the same formula for translational KE (Translational KE = (1/2) * m * V^2), we can equate the translational KE before and after the hill:
(1/2) * m * V^2 = (1/2) * m * V2^2
Canceling out the mass and solving for V2, we have:
V2 = V * sqrt(h2 / h1)
where h1 is the initial height (0 m) and h2 is the final height (2 m).
Substituting the values, we get:
V2 = 9 m/s * sqrt(2 / 0) = 9 m/s
So, the translational speed after the hill remains 9 m/s.
The rotational speed after the hill remains the same as well, as there is no external torque acting on the disk.
Therefore, the translational speed after the hill is 9 m/s and the rotational speed is still 18 rad/s.
The angular speed is w = V/R
I = moment of inertia = (1/2) M R^2
KE(rotational) = (1/2) I w^2
= (1/2)(1/2)M R^2(V/R)^2
= (1/4) M V^2
KE(ranslational) = (1/2) M V^2
Add the two KE types for total KE
Get to know these formulas. Others should not have to refresh your memory and do the work for oyu