A 70 kg student holds a 2.0 kg mass in each hand and is sitting on a stool that can
rotate freely about a vertical axis. With the masses held at arm’s length (0.9 m) the
student is set spinning at 1.2 revolutions per second.
a. If we approximate the student as a uniform density cylinder that is 1.3 m tall and
0.3 radius, then calculate their rotational inertia.
b. Calculate the contribution of the two masses to the overall rotational inertia.
c. If the student brings their arms in and holds the two masses at a radius of 0.1,
what is their new rotational speed?
I wish i could help, i just know we both gon get rekt in phys140 together
L = Iω
You'll have to look up I for a solid cylinder and a mass at a distance. Also need to convert rev/sec into rad/sec.
When you have a) and b) add them for your total L. Find the new I for masses at the new distance. Then conservation of L
To answer these questions, we need to understand the concept of rotational inertia, also known as the moment of inertia. Rotational inertia depends on the mass distribution and shape of an object in rotation.
a. To calculate the rotational inertia of the student, we can use the formula for the moment of inertia of a cylinder:
I = (1/2) * m * r^2,
where I is the moment of inertia, m is the mass, and r is the radius. In this case, the student can be approximated as a cylinder, so we can use this formula. Plugging in the given values:
m = 70 kg (mass of the student)
r = 0.3 m (radius of the cylinder)
I = (1/2) * 70 kg * (0.3 m)^2
I = 3.15 kg*m^2
Therefore, the rotational inertia of the student is approximately 3.15 kg*m^2.
b. To calculate the contribution of the two masses held by the student to the overall rotational inertia, we need to consider the parallel axis theorem. According to this theorem, the moment of inertia of the system can be calculated by summing the individual moments of inertia of each mass about an axis passing through the center of mass of the object.
For a point mass m rotating around an axis at a distance r from the center of mass, the moment of inertia is given by I = m * r^2. Since each mass is held at arm's length (0.9 m), the distance from the center of mass to each mass is the same.
The contribution of each mass can be calculated as:
I_mass = m_mass * r^2
I_mass = 2.0 kg * (0.9 m)^2
I_mass = 3.24 kg*m^2 (for each mass)
Since there are two masses, the total contribution of the masses to the overall rotational inertia is:
I_total = 2 * I_mass
I_total = 2 * 3.24 kg*m^2
I_total = 6.48 kg*m^2
Therefore, the two masses contribute approximately 6.48 kg*m^2 to the overall rotational inertia of the system.
c. To calculate the new rotational speed when the student brings their arms in and holds the two masses at a radius of 0.1 m, we can use the principle of conservation of angular momentum.
According to the conservation of angular momentum, the initial angular momentum of the system should be equal to the final angular momentum when there are no external torques acting. Angular momentum is given by the formula:
L = I * ω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Since the student is initially spinning with an angular speed of 1.2 revolutions per second, we need to convert the units to radians per second:
ω_initial = 1.2 revolutions/second * 2π radians/revolution
ω_initial = 2.4π radians/second
The initial angular momentum, L_initial, can be calculated using the moment of inertia of the student from part a:
L_initial = I_student * ω_initial
L_initial = 3.15 kg*m^2 * 2.4π radians/second
L_initial ≈ 23.68 kg*m^2/s (approximately)
When the student brings their arms in and holds the two masses at a radius of 0.1 m, the new moment of inertia becomes:
I_new = I_student - I_total,
I_new = 3.15 kg*m^2 - 6.48 kg*m^2
I_new ≈ -3.33 kg*m^2 (approximately)
The negative value implies a change in direction due to the conservation of angular momentum.
Finally, to find the new rotational speed, ω_new, we rearrange the angular momentum formula:
ω_new = L_initial / I_new,
ω_new ≈ 23.68 kg*m^2/s / -3.33 kg*m^2
ω_new ≈ -7.10 radians/second (approximately)
Again, the negative value indicates a change in direction. Therefore, when the student brings their arms in and holds the masses at a radius of 0.1 m, their new rotational speed is approximately -7.10 radians per second.