A circular disk of mass 0.2 kg and radius 24 cm, initially not rotating, slips down a thin spindle onto a turntable (disk) of mass 1.7 kg and the same radius, rotating freely at 3.4 rad/s.
Find the new angular velocity of the combination.
To find the new angular velocity of the combination, we can use the principle of conservation of angular momentum, which states that the total angular momentum before and after an event remains constant, provided no external torques act on the system.
The angular momentum of an object is given by the product of its moment of inertia (I) and its angular velocity (ω).
Before the disk slips onto the turntable, the angular momentum of the system is zero, since the disk is not rotating initially.
After the disk slips onto the turntable, the total angular momentum is the sum of the angular momentum of the disk and the angular momentum of the turntable:
L_total = L_disk + L_turntable
The moment of inertia of a disk is given by the formula:
I = (1/2) * m * r^2
Where m is the mass of the object and r is the radius.
The moment of inertia of the disk is:
I_disk = (1/2) * 0.2 kg * (0.24 m)^2 = 0.00576 kg * m^2
The moment of inertia of the turntable is:
I_turntable = (1/2) * 1.7 kg * (0.24 m)^2 = 0.09792 kg * m^2
The initial angular momentum of the system is zero, so we can write:
0 = L_disk + L_turntable
L_disk = -L_turntable
Using the formula for angular momentum, we can express the angular velocity of the disk and the turntable as:
L_disk = I_disk * ω_disk
L_turntable = I_turntable * ω_turntable
Since the disk is initially not rotating (ω_disk = 0), we have:
0 = 0.00576 kg * m^2 * 0 + 0.09792 kg * m^2 * ω_turntable
Simplifying the equation, we find:
ω_turntable = 0
This means that the angular velocity of the turntable remains unchanged after the disk slips onto it.
Therefore, the new angular velocity of the combination is also 3.4 rad/s.
To solve this problem, we can apply the principle of conservation of angular momentum. The angular momentum of an object is given by the product of its moment of inertia and angular velocity.
The angular momentum of the system before the disk is placed on the turntable is zero since the disk is not rotating.
After the disk is placed on the turntable, the angular momentum of the system is given by the sum of the angular momenta of the disk and the turntable.
The moment of inertia of a circular disk is given by the formula:
I = (1/2) * m * r^2
where m is the mass of the disk and r is the radius of the disk.
Given that the mass of the disk is 0.2 kg and the radius is 24 cm (0.24 m), the moment of inertia of the disk is:
I_disk = (1/2) * 0.2 kg * (0.24 m)^2
The moment of inertia of the turntable is given by the same formula, with the mass and radius of the turntable:
I_turntable = (1/2) * 1.7 kg * (0.24 m)^2
The initial angular momentum of the system is zero, so we can set the sum of the angular momenta of the disk and the turntable equal to zero:
I_disk * w_disk + I_turntable * w_turntable = 0
where w_disk is the angular velocity of the disk and w_turntable is the angular velocity of the turntable.
Substituting the expressions for the moment of inertia of the disk and turntable, we get:
(1/2) * 0.2 kg * (0.24 m)^2 * w_disk + (1/2) * 1.7 kg * (0.24 m)^2 * w_turntable = 0
Simplifying the equation, we have:
(0.864 * 0.2 kg * w_disk) + (0.864 * 1.7 kg * w_turntable) = 0
Now we can solve for w_turntable, the new angular velocity of the combination:
w_turntable = -((0.864 * 0.2 kg * w_disk) / (0.864 * 1.7 kg))
Given that the initial angular velocity of the turntable is 3.4 rad/s, we can substitute the value into the equation:
w_turntable = -((0.864 * 0.2 kg * 3.4 rad/s) / (0.864 * 1.7 kg))
Solving for w_turntable, we get:
w_turntable = -1.7 rad/s
Therefore, the new angular velocity of the combination is -1.7 rad/s.
a) I = 1/2 m r2
Iturn = 1/2 * 1.9 * 0.24^2= 0.05472 kg.m2
Idisc = 1/2 * 0.2 * 0.24^2 = 0.00576kg.m2
Itot = 0.05472 + 0.00576 = 0.162 kg.m2
angular momentum = 0.05472 * 3.4 = 0.186048 kg.m2/s
Itot * w = 0.186048
w = 0.186048 / 0.162
new angular velocity = 1.148 rad/s
b) change in KE = 1/2 If wf2 - 1/2 Ii wi2 = 1/2 ((0.162 * 1.148^2) - (0.186048 * 3.4^2))
= -0.968 J
c) T = I * alpha
= 0.162 * [3.4 - 1.148 / 3.4]
constant torque = 0.107 N.m