1. A centrifuge in a medical laboratory rotates at an angular speed of 3650 rev/min. When switched off, it rotates through 50.0 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge.

2. A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

3. A potter's wheel moves from rest to an angular speed of 0.19 rev/s in 35 s. Find its angular acceleration in radians per second per second.

1.Divide the initial ANGULAR velocity in radians/sec by the time required to stop.

The initial angular velocity is 3650 rev/min*(2 pi rad/rev)/60 (s/min) = 382.2 rad/s. The time required to stop is
(Number of revolutions)/(average speed) = 50 rev/1825 rev/min = .0274 min = 1.644 s.
angular acceleration = -382.2 rad/s/1.644 s = ? rad/s^2
The minus sign is there becasue it is slowing down.

2. Impulse = change in momentum
Don't forget that the sign of the momentum changes. You will need to calculate the velocity sqrt(2gH) when the ball hits the floor. It will rebound with a lower velocity that you also need to calculate. You can use the same sqrt(2gH) formula with a lower value for H.

3. Use method similar to (1), but the initial velocity is in rev/s this time, not rev/min

1. To find the constant angular acceleration, we can use the following formula:

ωf = ωi + αt

where ωf is the final angular speed, ωi is the initial angular speed, α is the angular acceleration, and t is the time.

Given:
ωi = 3650 rev/min
ωf = 0 rev/min (since it comes to rest)
t = 50.0 revolutions

First, we need to convert the initial and final angular speeds to radians per second:
ωi = 3650 rev/min * (2π rad/rev) * (1 min / 60 s) = 381.7 rad/s
ωf = 0 rev/min * (2π rad/rev) * (1 min / 60 s) = 0 rad/s

Next, we'll convert the time to seconds:
t = 50.0 revolutions * (2π rad/rev) * (1 / ωi) = 51.9 s

Now, we can substitute the values into the formula and solve for α:
0 rad/s = 381.7 rad/s + α * 51.9 s

Rearranging the equation, we have:
α = -381.7 rad/s / 51.9 s

Calculating this, we get:
α = -7.35 rad/s^2

Therefore, the constant angular acceleration of the centrifuge is -7.35 rad/s^2.

2. To find the impulse given to the ball by the floor, we can use the principle of conservation of linear momentum:

Impulse = Change in momentum

The momentum of an object is given by the formula:

p = mv

where p is the momentum, m is the mass, and v is the velocity.

Given:
m = 0.120 kg
Initial height (h1) = 1.25 m
Final height (h2) = 0.600 m

The initial velocity can be calculated using the equation of motion:

v^2 = u^2 + 2gh

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height.

Plugging in the values:
v^2 = 0 + 2 * 9.8 m/s^2 * (h1 - h2)
= 2 * 9.8 m/s^2 * (1.25 m - 0.600 m)
= 19.6 m^2/s^2 * 0.650 m
= 12.74 m^2/s^2

Taking the square root of both sides:
v = √(12.74 m^2/s^2)
= 3.57 m/s (approx)

The initial momentum is then:
p1 = mv
= 0.120 kg * 0 m/s
= 0 kg*m/s

The final momentum is:
p2 = mv
= 0.120 kg * 3.57 m/s
= 0.4284 kg*m/s

Therefore, the impulse given to the ball by the floor is the change in momentum:
Impulse = p2 - p1
= 0.4284 kg*m/s - 0 kg*m/s
= 0.4284 kg*m/s

So, the impulse given to the ball by the floor is 0.4284 kg*m/s.

3. To find the angular acceleration in radians per second per second, we can use the formula:

α = (ωf - ωi) / t

where α is the angular acceleration, ωf is the final angular speed, ωi is the initial angular speed, and t is the time.

Given:
ωi = 0 rev/s (since it starts from rest)
ωf = 0.19 rev/s
t = 35 s

First, we need to convert the angular speeds to radians per second:
ωi = 0 rev/s * (2π rad/rev) = 0 rad/s
ωf = 0.19 rev/s * (2π rad/rev) = 1.19 rad/s

Now, we can substitute the values into the formula and solve for α:
α = (1.19 rad/s - 0 rad/s) / 35 s

Calculating this, we get:
α = 0.034 radians per second per second

Therefore, the angular acceleration of the potter's wheel is 0.034 radians per second per second.

1. To find the constant angular acceleration of the centrifuge, we can use the equation of motion for rotational motion:

ω^2 = ω0^2 + 2αθ

where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and θ is the angle rotated.

Given:
ω = 0 (since the centrifuge comes to rest)
ω0 = 3650 rev/min
θ = 50.0 revolutions = 50.0 * 2π radians

First, let's convert ω0 from rev/min to radians/s:
ω0 = (3650 rev/min) * (2π rad/rev) * (1 min/60 s) = 381.08 rad/s

Now we can plug in the values into the equation:

0^2 = (381.08)^2 + 2α(50.0 * 2π)

Simplifying the equation:

0 = 145637.3664 + 628.32πα

Now we can solve for α:

628.32πα = - 145637.3664

α = - 145637.3664 / (628.32π)

Calculating the value:

α ≈ -7.33 rad/s^2

Therefore, the constant angular acceleration of the centrifuge is approximately -7.33 rad/s^2.

2. To find the impulse given to the ball by the floor, we can use the impulse-momentum principle, which states that the impulse exerted on an object is equal to the change in its momentum:

Impulse = Change in momentum

We can calculate the change in momentum by using the equation:

Change in momentum = m(v - u)

where m is the mass of the ball, v is the final velocity, and u is the initial velocity.

Given:
m = 0.120 kg
v (after rebound) = 0 m/s (as it reaches the maximum height and comes to rest momentarily)
u (before rebound) = ? (unknown)

To find u, we can use the conservation of mechanical energy:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

At the topmost point (maximum height), the ball is momentarily at rest, so the final kinetic energy is 0. The final potential energy is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Initial kinetic energy = (1/2) mu^2
Initial potential energy = mgh

Given:
h1 (initial height) = 1.25 m
h2 (final height) = 0.600 m
g (acceleration due to gravity) = 9.8 m/s^2

Using the conservation of mechanical energy equation:

(1/2) mu^2 + mgh1 = 0 + mgh2

(1/2) u^2 + gh1 = gh2

Simplifying the equation:

(1/2) u^2 = g(h2 - h1)

Substituting the values:

(1/2) u^2 = 9.8(0.600 - 1.25)

(1/2) u^2 ≈ -4.9

Now we can solve for u:

u^2 ≈ -9.8

Taking the square root of both sides:

u ≈ ±3.13 m/s

Since the ball is dropped from rest, the initial velocity (u) should be negative, so we take:

u ≈ -3.13 m/s

Now we can calculate the impulse:

Impulse = m(v - u)

Impulse = (0.120 kg)(0 - (-3.13 m/s)) = 0.120 kg * 3.13 m/s = 0.376 Ns

Therefore, the impulse given to the ball by the floor is approximately 0.376 Ns.

3. To find the angular acceleration of the potter's wheel, we can use the equation of motion for rotational motion:

ω = ω0 + αt

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

Given:
ω = 0.19 rev/s
ω0 = 0 (since the wheel starts from rest)
t = 35 s

We'll convert ω from rev/s to radians/s:

ω = (0.19 rev/s) * (2π rad/rev)

ω = 0.19 * 2π rad/s

Now we can plug in the values into the equation:

0.19 * 2π = 0 + α * 35

Simplifying the equation:

0.38π = 35α

Now we can solve for α:

α = 0.38π / 35

Calculating the value:

α ≈ 0.034 rad/s^2

Therefore, the angular acceleration of the potter's wheel is approximately 0.034 rad/s^2.