Two uniform solid spheres have the same mass of 3 kg, but one has a radius of 0.25 m while the other has a radius of 0.8 m. Each can rotate about an axis through its center.
(a) What is the magnitude τ of the torque required to bring the smaller sphere from rest to an angular speed of 300 rad/s in 15 s?
N·m
(b) What is the magnitude F of the force that must be applied tangentially at the sphere's equator to provide that torque?
N
(c) What is the corresponding value of τ for the larger sphere?
N·m
(d) What is the corresponding value of F for the larger sphere?
N
I sphere = (2/5) mR^2
see http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html
I1 = (2/5)3 (.25)^2 = .075
alpha = 300/15 = 20 radians/second^2
Torque = I alpha = .075*20 = 1.5 Nm
F1 = torque/R = 1.5/.25 = 6 N
now big sphere
I2 = (2/5)3 (.8)^2 = .768
torque = I alpha = .768*20 = 15.36 Nm
F2 = 15.36/.8 = 19.2 N
To solve these problems, we will use the formulas for torque and moment of inertia.
The torque, τ, acting on an object is given by the formula:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The moment of inertia, I, for a solid sphere rotating about its axis is given by the formula:
I = (2/5) * m * r^2
where m is the mass of the sphere and r is the radius.
(a) To find the torque required to bring the smaller sphere from rest to an angular speed of 300 rad/s in 15 s, we need to calculate the moment of inertia, I, and the angular acceleration, α.
Given:
Mass of the smaller sphere, m = 3 kg
Radius of the smaller sphere, r = 0.25 m
Angular speed, ω = 300 rad/s
Time, t = 15 s
First, we can calculate the angular acceleration using the formula:
α = ω/t = 300 rad/s / 15 s
α = 20 rad/s^2
Next, we can calculate the moment of inertia using the formula:
I = (2/5) * m * r^2 = (2/5) * 3 kg * (0.25 m)^2
I = 0.3 kg·m^2
Finally, we can find the torque using the formula:
τ = Iα = 0.3 kg·m^2 * 20 rad/s^2
τ = 6 N·m
Therefore, the magnitude of the torque required to bring the smaller sphere from rest to an angular speed of 300 rad/s in 15 s is 6 N·m.
(b) To find the magnitude of the force, F, that must be applied tangentially at the sphere's equator to provide that torque, we can use the formula:
τ = Fr
where F is the force applied and r is the radius of the sphere.
Rearranging the formula, we get:
F = τ / r = 6 N·m / 0.25 m
F = 24 N
Therefore, the magnitude of the force that must be applied tangentially at the sphere's equator is 24 N.
(c) To find the corresponding value of τ for the larger sphere, we can reuse the formula for moment of inertia:
I = (2/5) * m * r^2
Given:
Mass of the larger sphere, m = 3 kg
Radius of the larger sphere, r = 0.8 m
Using the formula, we get:
I = (2/5) * 3 kg * (0.8 m)^2
I = 3.84 kg·m^2
Therefore, the moment of inertia for the larger sphere is 3.84 kg·m^2.
(d) To find the corresponding value of F for the larger sphere, we can use the formula:
F = τ / r
Given:
Torque, τ = 6 N·m
Radius of the larger sphere, r = 0.8 m
Using the formula, we get:
F = 6 N·m / 0.8 m
F = 7.5 N
Therefore, the magnitude of the force that must be applied tangentially at the larger sphere's equator is 7.5 N.
To solve these problems, we will be using the formulas for torque and rotational inertia.
The torque (τ) required to bring a sphere from rest to an angular speed can be calculated using the formula:
τ = Iα
where τ is the torque, I is the rotational inertia, and α is the angular acceleration.
The rotational inertia (I) of a solid sphere can be calculated using the formula:
I = (2/5) * m * r^2
where m is the mass of the sphere and r is the radius of the sphere.
(a) To find the magnitude of the torque required to bring the smaller sphere from rest to an angular speed of 300 rad/s in 15 s:
Given:
Mass (m) = 3 kg
Radius (r) = 0.25 m
Angular speed (ω) = 300 rad/s
Time (t) = 15 s
Using the formula for rotational inertia, we can calculate the rotational inertia of the smaller sphere:
I = (2/5) * m * r^2
= (2/5) * 3 kg * (0.25 m)^2
= 0.3 kg·m^2
To find the angular acceleration (α), we can use the formula:
α = (ω - ω0) / t
= (300 rad/s - 0 rad/s) / 15 s
= 20 rad/s^2
Now, we can calculate the magnitude of the torque (τ):
τ = Iα
= 0.3 kg·m^2 * 20 rad/s^2
= 6 N·m
Therefore, the magnitude of the torque required is 6 N·m.
(b) To find the magnitude of the force (F) that must be applied tangentially at the sphere's equator to provide that torque:
We can use the formula:
τ = rFsinθ
where r is the radius of the sphere and θ is the angle between the force and the lever arm.
Since the torque is equal to rFsinθ, and we want to find the magnitude of the force (F), we can rearrange the formula as follows:
F = τ / (r * sinθ)
In this case, the sphere is rotating about an axis through its center, so the angle (θ) is 90 degrees.
Using this information, we can calculate the magnitude of the force (F):
F = τ / (r * sinθ)
= 6 N·m / (0.25 m * sin(90°))
= 24 N
Therefore, the magnitude of the force required is 24 N.
(c) To find the corresponding value of τ for the larger sphere:
We know that both spheres have the same mass of 3 kg, but the radius (r) of the larger sphere is 0.8 m.
Using the rotational inertia formula, we can calculate the rotational inertia of the larger sphere:
I = (2/5) * m * r^2
= (2/5) * 3 kg * (0.8 m)^2
= 3.84 kg·m^2
Therefore, the corresponding value of the torque (τ) for the larger sphere is 3.84 N·m.
(d) To find the corresponding value of F for the larger sphere:
Using the same formula as in part (b):
F = τ / (r * sinθ)
= 3.84 N·m / (0.8 m * sin(90°))
= 4.8 N
Therefore, the corresponding value of the force required for the larger sphere is 4.8 N.