A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 5 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down?

=> Using the rotational analogue of the common equation of motion 2as = v^2 - u^2 , we get:

2 x (Ang. Acc) x (Ang. Distance) = (Final Angular Vel.)^2 - (Initial Angular Vel.)^2

=> Since it comes to a stop, the Final Ang. Vel. is zero. The intial is given as 45 rev per minute, which is 3/4 revs per second. The Ang. Distance is given as 5 complete turns, which is 10π rad.

So, using the above formula:

Ang. Acc. = - (3/4)^2/2(10π)
= - 9/320π

To solve this problem, we need to use the equations of rotational motion. The first thing we need to do is convert the given quantities into the appropriate units.

1 revolution = 2π radians.
45 revolutions per minute = (2π * 45) radians per minute.

Now we need to find the angular acceleration. We can use the formula:

θ = 0.5 * α * t^2,

where θ is the angular displacement, α is the angular acceleration, and t is the time.

From the problem statement, we know that the record made 5 complete turns, which is equivalent to an angular displacement of 10π radians (since 1 complete turn = 2π radians).

Substituting the given values into the equation, we get:

10π = 0.5 * α * t^2.

Now we need to find the time it took for the record to slow down and come to rest. The formula is given by:

ω = ω0 + α * t,

where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.

Since the record is rotating at 45 revolutions per minute before coming to rest, the initial angular velocity, ω0, is (2π * 45) radians per minute.

When the record player is switched off, it comes to rest, so the final angular velocity, ω, is 0.

Substituting these values into the equation, we get:

0 = (2π * 45) + α * t.

Now we have a system of equations:

10π = 0.5 * α * t^2,
0 = (2π * 45) + α * t.

We can solve this system of equations simultaneously to find the value of α, the angular acceleration.

First, let's isolate t from the second equation:

α * t = - (2π * 45).

Now substitute this value of α * t into the first equation:

10π = 0.5 * (-2π * 45)^2.

Simplifying further:

10π = 0.5 * 4 * (π^2) * (45^2).

Dividing both sides by 10 * π:

1 = 1800 * 45 * π, or
1 = 81000 * π.

Dividing both sides by 81000:

1/81000 = π.

Therefore, the magnitude of the angular acceleration, α, is 1/81000 radians per second squared.

To find the magnitude of the angular acceleration, we need to determine the total angular displacement of the record while it was slowing down.

First, we should convert the given rotation rate of the record player from revolutions per minute (rpm) to radians per second (rad/s).

1 revolution = 2π radians
45 rpm = 45 * 2π rad/1 min = 45 * 2π / 60 rad/s = 3π/2 rad/s

Now, we can calculate the total angular displacement of the record during its deceleration.

Total angular displacement (θ) = (number of turns) * (360° or 2π radians per turn)

Given that the record makes 5 complete turns, the total angular displacement is:

θ = 5 * 2π = 10π radians

Next, we need to use the formula for angular displacement that involves angular acceleration to find the magnitude of the angular acceleration (α):

θ = (initial angular velocity * time) + (0.5 * α * time^2)

Since the record comes to rest, the initial angular velocity is 3π/2 rad/s, and the time it takes to slow down is unknown. Instead of the time, we can use the number of turns (n) and the rotation rate (ω), where ω = α * time.

θ = (ω * n) + (0.5 * α * n^2)

10π = (3π/2 * n) + (0.5 * α * n^2)

Simplifying the equation:

10 = (3/2 * n) + (0.5 * α * n^2)

Now, since we know that the record makes 5 complete turns (n = 5), we can substitute this into the equation:

10 = (3/2 * 5) + (0.5 * α * 5^2)

Now we can solve for α:

10 = (15/2) + (0.5 * α * 25)
10 - 15/2 = 12.5 * α
20 - 15/2 = 12.5 * α
20 - 7.5 = 12.5 * α
12.5 = 12.5 * α
α = 1 rad/s^2

Therefore, the magnitude of the angular acceleration of the record as it slowed down was 1 rad/s^2.