A uniform solid disk of radius 1.7 m and mass 81.3 kg is free to rotate on a frictionless pivot through a point on its rim.
If the disk is released from rest in the position shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? The acceleration of gravity is 9.8 m/s^2.
The figure shows a solid circle, with a pivot point on the leftmost point of the rim. The dashed circle is basically rotated so that the pivot point is at the very top.
The center of mass goes down by the radius r
the change in potential energy is m g r
that is the kinetic energy
(1/2) I w^2
I is the moment of inertia about the pivot point which is the moment of inertia about the center + m r^2
the velocity v = w r
To find the speed of the center of mass when the disk reaches the dashed circle position, we need to consider the conservation of angular momentum.
The angular momentum of the disk is given by the equation:
L = Iω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a uniform solid disk rotating about an axis perpendicular to its plane through its center is given by:
I = (1/2)mr^2,
where m is the mass of the disk and r is the radius.
In this case, the disk starts from rest, so the initial angular velocity is zero (ω_initial = 0). At the final position indicated by the dashed circle, the angular velocity is ω_final.
Since angular momentum is conserved, we have:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Substituting the expressions for moment of inertia and angular velocities:
(1/2)mr^2 * 0 = (1/2)mr^2 * ω_final
Simplifying the equation:
0 = ω_final
This means that the angular velocity at the final position is zero.
However, the speed of the center of mass is related to the angular velocity by the equation:
v = r * ω
where v is the speed of the center of mass and r is the radius.
Since ω_final = 0, the speed of the center of mass is also zero:
v = r * ω_final = 0.
Therefore, the speed of the center of mass when the disk reaches the position indicated by the dashed circle is zero.
To find the speed of the center of mass when the disk reaches the position indicated by the dashed circle, we can use the principle of conservation of mechanical energy.
First, let's denote the initial position of the solid circle as point A (at rest) and the final position of the dashed circle as point B.
The mechanical energy of the disk is conserved since there are no external forces acting on it. Therefore, the sum of the kinetic energy and the potential energy at point A is equal to the sum of the kinetic energy and the potential energy at point B.
At point A, the disk is at rest, so its kinetic energy is zero. The potential energy at point A is given by the gravitational potential energy:
PE_A = m * g * h_A
where m is the mass of the disk, g is the acceleration due to gravity, and h_A is the height of point A above some reference point.
At point B, the disk has rotational kinetic energy due to its angular velocity, and the potential energy at point B is given by the height of point B:
KE_B + PE_B = 1/2 * I * omega^2 + m * g * h_B
where I is the moment of inertia of the disk, omega is its angular velocity, and h_B is the height of point B above the reference point.
The moment of inertia of a uniform solid disk rotating around an axis perpendicular to its plane through its center is given by:
I = (1/2) * m * r^2
where r is the radius of the disk.
Since the disk is released from rest, its angular velocity at point B is the same as its angular acceleration. The angular acceleration can be determined using torque equations, but for this problem, we can use the equation:
vf = sqrt(2 * a * s)
where vf is the final velocity, a is the acceleration, and s is the distance traveled. In our case, the final velocity is the angular velocity at point B, the acceleration is the angular acceleration, and the distance traveled is the angular displacement of the disk.
Since the distance traveled is the entire circumference of the disk (2πr), the angular displacement is 2π radians.
Applying this equation:
vf = sqrt(2 * a * s)
vf = sqrt(2 * a * 2πr)
vf = sqrt(4πar)
Now, since the angular displacement is 2π radians, the final angular velocity in terms of angular acceleration can be calculated as:
vf = ω * r
sqrt(4πar) = ω * r
ω = sqrt(4πa)
Using this expression for ω, we can rewrite the equation for the kinetic energy at point B:
KE_B + PE_B = 1/2 * I * ω^2 + m * g * h_B
KE_B + PE_B = 1/2 * (1/2) * m * r^2 * (4πa) + m * g * h_B
KE_B + PE_B = 1/4 * m * r^2 * (4πa) + m * g * h_B
KE_B + m * g * h_B = m * r^2 * πa + m * g * h_B
KE_B = m * r^2 * πa
Since the mechanical energy is conserved, we can equate the potential energy terms:
PE_A = PE_B
m * g * h_A = m * g * h_B
Rearranging for h_B:
h_B = h_A
Now we can substitute the expression for m * g * h_B into the equation for KE_B:
KE_B = m * r^2 * πa
Finally, the speed of the center of mass is given by the linear velocity of a point on the rim of the disk, which can be calculated as:
v = ω * r
Substituting the expression for ω obtained earlier:
v = sqrt(4πa) * r
Now we have the expression for the speed of the center of mass of the disk when it reaches the position indicated by the dashed circle.