Final answer:
The angular momentum of the proton about the center at its maximum speed is approximately 4.17 x 10^-18 kg m^2/s. The torque on the proton about the center as it accelerates to maximum speed is approximately 4.17 x 10^-17 Nm.
Step-by-step explanation:
To find the angular momentum of the proton about the center at maximum speed, we can use the formula:
L = mvr
where L is the angular momentum, m is the mass of the proton, v is the velocity, and r is the radius of the circular path.
Given that the mass of a proton is approximately 1.67 x 10^-27 kg, the velocity is 5.0 x 10^6 m/s, and the radius is 0.5 km (which is 0.5 x 10^3 m), we can calculate:
L = (1.67 x 10^-27 kg)(5.0 x 10^6 m/s)(0.5 x 10^3 m)
L = 4.17 x 10^-18 kg m^2/s
The angular momentum of the proton about the center at its maximum speed is approximately 4.17 x 10^-18 kg m^2/s.
The torque on the proton as it accelerates to maximum speed can be calculated using the equation:
T = Iα
where T is the torque, I is the moment of inertia, and α is the angular acceleration.
Since the proton follows a circular path, we can rewrite the moment of inertia as:
I = mr^2
Substituting the values given, we have:
T = (1.67 x 10^-27 kg)(0.5 x 10^3 m)^2α
The angular acceleration α can be calculated using the equation:
α = Δv / Δt
where Δv is the change in velocity and Δt is the change in time.
Given that the proton accelerates from 0 m/s to 5.0 x 10^6 m/s in 0.01 s, we can calculate:
α = (5.0 x 10^6 m/s - 0 m/s) / (0.01 s)
α = 5.0 x 10^8 m/s^2
Substituting this value back into the torque equation, we get:
T = (1.67 x 10^-27 kg)(0.5 x 10^3 m)^2(5.0 x 10^8 m/s^2)
T = 4.17 x 10^-17 Nm
The torque on the proton about the center as it accelerates to maximum speed is approximately 4.17 x 10^-17 Nm.