Gyroscope homework question

[rytagi17]

A gyroscope flywheel of radius 3.33 cm is accelerated from rest at 13.4 rad/s2 until its angular speed is 2760 rev/min.
(a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?
m/s2
(b) What is the radial acceleration of this point when the flywheel is spinning at full speed?
m/s2
(c) Through what distance does a point on the rim move during the spin-up?
m

i've been trying to answer these for hours and still have no clue. Help is appreciated.

 

[D H]

We don't do your homework problems for you here. Instead, we help you come to the solution yourself. When you posted this problem you were presented with a template. It would have helped if you had followed it. What are the relevant equations in this problem? What work have you done to solve the problem?

 

[Moetasim]

I would recommend using the following equations to solve this problem:

(a) The tangential acceleration of a point on the rim of the flywheel can be calculated using the equation a = rα, where a is the tangential acceleration, r is the radius of the flywheel, and α is the angular acceleration. Plugging in the values given in the question, we get:

a = (0.0333 m)(13.4 rad/s^2) = 0.447 m/s^2

Therefore, the tangential acceleration of a point on the rim of the flywheel during the spin-up process is 0.447 m/s^2.

(b) The radial acceleration of a point on the rim can be calculated using the equation ar = v^2/r, where ar is the radial acceleration, v is the tangential velocity, and r is the radius. At full speed, the tangential velocity can be calculated by converting the angular speed from revolutions per minute (rpm) to radians per second (rad/s) and then multiplying it by the radius:

v = (2760 rpm)(2π rad/rev)(1 min/60 s)(0.0333 m) = 9.17 m/s

Plugging in this value and the given radius into the equation, we get:

ar = (9.17 m/s)^2/(0.0333 m) = 2520 m/s^2

Therefore, the radial acceleration of a point on the rim when the flywheel is spinning at full speed is 2520 m/s^2.

(c) To calculate the distance traveled by a point on the rim during the spin-up process, we can use the equation d = 1/2αt^2, where d is the distance traveled, α is the angular acceleration, and t is the time. To find the time, we can use the equation ω = ω0 + αt, where ω is the final angular speed, ω0 is the initial angular speed (which is 0 in this case), and t is the time. Rearranging this equation, we get:

t = (ω - ω0)/α = (2760 rpm)(2π rad/rev)(1 min/60 s)/(13.4 rad/s^2) = 6.52 s

Plugging this value and the given angular acceleration into the distance equation, we get