How to solve?
2. A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 10.0 m) and spun at a constant angular velocity of 16.3 rpm. Answer the following:
a. What is the angular velocity of the centrifuge in rad/s?
b. What is the linear velocity of the astronaut at the outer edge of the centrifuge?
c. What is the centripetal acceleration of the astronaut at the end of the centrifuge?
d. How many g’s does the astronaut experience?
e. What is the centripetal force and net torque experienced by the astronaut? Give magnitudes and directions.
w = omega =
16.3revs/min * 2 pi rad/rev *(1 min/60 s)
= 1.71 radians/s
v = omega r = 1.71 * 10 = 17.1 m/s
Ac = v^2/r = 29.1 m/s^2
29.1/9.81 = 2.97 gs
F = m a = 70 * 2.97 = 208 Newtons toward center
There is no angular acceleration so no torque as far as I know.
Well, this astronaut is really spinning into action! Let's break down the solutions for each part of the problem.
a. To convert the angular velocity from rpm to rad/s, we use the conversion factor: 1 rpm = 2π rad/min. So, first, we figure out how many rad/min the centrifuge is spinning: 16.3 rpm * 2π rad/min = ? rad/min. But we need the result in rad/s, not rad/min. So, we convert the result to rad/s: ? rad/min * 1 min/60 s = ? rad/s. Don't forget to plug in the numbers!
b. To find the linear velocity of the astronaut at the outer edge of the centrifuge, we use the formula v = r * ω, where v is the linear velocity, r is the radius, and ω is the angular velocity. Simply plug in the given values to calculate the linear velocity.
c. For the centripetal acceleration, we use the formula a = r * ω^2. Plug in the values you have to calculate the centripetal acceleration.
d. Oh, "g's"! It's like being at a carnival ride, isn't it? Here, "g" represents the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2. To find how many g's the astronaut experiences, we divide the centripetal acceleration by g.
e. To calculate the centripetal force, we use the formula F = m * a_c, where F is the centripetal force, m is the mass of the astronaut, and a_c is the centripetal acceleration. For the net torque, we need the formula τ = r * F, where τ is the torque, r is the radius, and F is the centripetal force. Plug in the given values to calculate both the force and the torque, don't forget the directions!
Remember, solving physics problems is not rocket science... or wait, it actually is rocket science. But don't worry, you can handle it!
To solve this problem, we will use the following formulas:
1. Angular velocity (ω) in rad/s = 2π × (angular velocity in rpm) / 60
2. Linear velocity (v) = radius (r) × angular velocity (ω)
3. Centripetal acceleration (a) = (linear velocity (v))^2 / radius (r)
4. g-force (g) = centripetal acceleration (a) / 9.8 m/s²
5. Centripetal force (Fcp) = mass (m) × centripetal acceleration (a)
6. Net torque = 0 (because there is no change in angular velocity)
Now let's solve the problem step by step:
a. Angular velocity of the centrifuge in rad/s:
ω = 2π × (16.3 rpm) / 60
= 2π × 16.3 / 60
≈ 1.714 rad/s
b. Linear velocity of the astronaut at the outer edge of the centrifuge:
v = radius × angular velocity
= 10.0 m × 1.714 rad/s
≈ 17.14 m/s
c. Centripetal acceleration of the astronaut at the end of the centrifuge:
a = (v)^2 / r
= (17.14 m/s)^2 / 10.0 m
≈ 29.339 m/s²
d. g-force experienced by the astronaut:
g = a / 9.8 m/s²
≈ 29.339 m/s² / 9.8 m/s²
≈ 2.995 g (approximately 3 g's)
e. Centripetal force and net torque experienced by the astronaut:
Centripetal force (Fcp) = mass × centripetal acceleration
Fcp = 70.0 kg × 29.339 m/s²
≈ 2053.73 N
Net Torque = 0 (since there is no change in angular velocity)
Therefore, the centripetal force experienced by the astronaut is approximately 2053.73 N, directed towards the center of the centrifuge. And the net torque experienced by the astronaut is zero.
To solve this problem, we can use the following formulas:
a. Angular velocity (in radians per second) is given by the formula:
ω = 2πf
Where ω is the angular velocity and f is the frequency (in Hz). In this case, the frequency is given in rpm (revolutions per minute), so we need to convert it to Hz before using the formula.
To convert rpm to Hz, we can use the formula:
f = rpm / 60
Now, we can substitute the values and calculate the angular velocity (in rad/s).
b. The linear velocity at the outer edge of the centrifuge can be calculated using the formula:
v = rω
Where v is the linear velocity, r is the radius, and ω is the angular velocity (in rad/s). We already calculated ω in the previous step, so we can substitute the values to find the linear velocity.
c. The centripetal acceleration can be calculated using the formula:
a = rω^2
Where a is the centripetal acceleration, r is the radius, and ω is the angular velocity (in rad/s). We have the values for r and ω, so we can substitute them to find the centripetal acceleration.
d. The acceleration in terms of "g" can be calculated by dividing the centripetal acceleration by the acceleration due to gravity (g ≈ 9.8 m/s^2). This will give you the number of "g's" the astronaut experiences.
e. To calculate the centripetal force, we can use the formula:
Fc = ma
Where Fc is the centripetal force, m is the mass of the astronaut, and a is the centripetal acceleration. We have the values for m and a, so we can substitute them to find the centripetal force.
The net torque experienced by the astronaut can be calculated using the formula:
τ = Iα
Where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, α is assumed to be zero since the astronaut is rotating at a constant angular velocity. Therefore, the net torque experienced by the astronaut would be zero.
Now, let's calculate the answers to each question using the given values:
a. The angular velocity in rad/s can be calculated as:
f = 16.3 rpm / 60 ≈ 0.272 Hz
ω = 2πf ≈ 2π * 0.272 ≈ 1.71 rad/s
b. The linear velocity at the outer edge of the centrifuge can be calculated as:
v = rω = 10.0 m * 1.71 rad/s ≈ 17.1 m/s
c. The centripetal acceleration at the end of the centrifuge can be calculated as:
a = rω^2 = 10.0 m * (1.71 rad/s)^2 ≈ 29.48 m/s^2
d. To calculate the number of "g's" the astronaut experiences, we divide the centripetal acceleration by the acceleration due to gravity:
g = 9.8 m/s^2
Number of g's = a / g ≈ 29.48 m/s^2 / 9.8 m/s^2 ≈ 3.01 g
e. The centripetal force experienced by the astronaut can be calculated as:
Fc = ma = 70.0 kg * 29.48 m/s^2 ≈ 2063.6 N (directed towards the center of rotation)
The net torque experienced by the astronaut is assumed to be zero since there is no angular acceleration.
Therefore, the answers to the given questions are:
a. The angular velocity of the centrifuge is approximately 1.71 rad/s.
b. The linear velocity of the astronaut at the outer edge of the centrifuge is approximately 17.1 m/s.
c. The centripetal acceleration of the astronaut at the end of the centrifuge is approximately 29.48 m/s^2.
d. The astronaut experiences approximately 3.01 g's.
e. The centripetal force experienced by the astronaut is approximately 2063.6 N (directed towards the center of rotation), and the net torque is assumed to be zero.