A quarter on a spinning vinyl record has a centripetal acceleration of 31.8 m/s2. What is the acceleration of the quarter if it is placed twice as far from the centre of the vinyl record?
To find the acceleration of the quarter when it is placed twice as far from the center of the vinyl record, we can use the concept of centripetal acceleration.
The formula for centripetal acceleration is:
a = (v^2) / r
Where:
a = centripetal acceleration
v = tangential velocity
r = radius of the circular path
In this case, we are given the centripetal acceleration (31.8 m/s^2) for the original position of the quarter on the vinyl record. Since the centripetal acceleration does not depend on the mass of the object, it will remain constant regardless of the radius.
Let's assume the original radius is 'r'. Therefore, the original tangential velocity, v, can be calculated using the centripetal acceleration formula:
31.8 = (v^2) / r
Now, when we place the quarter twice as far from the center, the new radius becomes '2r'. We want to find the new acceleration, 'a_new'.
Using the same centripetal acceleration formula, we can rewrite it using the new parameters:
a_new = (v_new^2) / (2r)
However, we have to consider that the tangential velocity, v_new, will be different from the original tangential velocity, v. To find v_new, we need to use the conservation of angular momentum.
Angular momentum is given by the equation:
L = Iω
Where:
L = angular momentum
I = moment of inertia
ω = angular velocity
Since there are no external torques acting on the system, the angular momentum remains constant.
Initially:
L_initial = I_initial * ω_initial
Finally:
L_final = I_final * ω_final
Since the moment of inertia is proportional to r^2, we can write:
L_initial = I_initial * ω_initial = r^2 * ω_initial
L_final = I_final * ω_final = (2r)^2 * ω_final = 4r^2 * ω_final
Since angular momentum is conserved, we can equate the initial and final angular momentum:
L_initial = L_final
r^2 * ω_initial = 4r^2 * ω_final
Cancelling out r^2 from both sides, we get:
ω_initial = 4ω_final
So the initial angular velocity is 4 times the final angular velocity.
Now, the tangential velocity, v, is related to the angular velocity, ω, by the formula:
v = r * ω
For the initial position:
v_initial = r * ω_initial
v_initial = r * (4ω_final)
For the final position:
v_final = (2r) * ω_final
We can substitute these values back into the formula for centripetal acceleration to find the new acceleration (a_new):
a_new = (v_final^2) / (2r)
a_new = ((2r * ω_final)^2) / (2r)
a_new = (4r^2 * ω_final^2) / (2r)
a_new = 2r * ω_final^2
But we know that:
ω_initial = 4ω_final
So we can substitute ω_initial/4 for ω_final in the equation for a_new:
a_new = 2r * (ω_initial/4)^2
a_new = 2r * (1/16) * (ω_initial^2)
a_new = (r/8) * (ω_initial^2)
Finally, using the initial centripetal acceleration formula (31.8 = (v^2) / r), we can solve for v^2:
31.8 = (v_initial^2) / r
v_initial^2 = 31.8 * r
Substituting v_initial^2 in the equation for a_new:
a_new = (r/8) * (31.8 * r)
a_new = (r/8) * 31.8 * r
a_new = 3.975 * r^2
Therefore, the acceleration of the quarter when it is placed twice as far from the center of the vinyl record is 3.975 times greater than the initial centripetal acceleration.