Torque=momentInertia*deacceleration
The average speed during deacceleration is 1.33/2, so the time to stop is displacement/avg velocity,or .75*2PI/(1.33/2)
deacceleration = wi/time
so you can find torque.
(a) Find the average torque exerted on the wheel given that it is a disk of radius 0.76 m and mass 6.4 kg.
? N·m
sorry, I'm having converting issues...I don't see the link between theta, omega, moment of inirtia, torque...etc...
The average speed during deacceleration is 1.33/2, so the time to stop is displacement/avg velocity,or .75*2PI/(1.33/2)
deacceleration = wi/time
so you can find torque.
The moment of inertia (I) represents how the mass of an object is distributed around its axis of rotation. For a disk, the moment of inertia is given by the equation:
I = (1/2) * m * r^2
where m is the mass of the disk and r is the radius. Let's substitute the given values into this equation:
I = (1/2) * 6.4 kg * (0.76 m)^2 = 1.463 kg·m^2
The angular acceleration (α) is the rate at which angular velocity changes. It can be determined using the equation:
α = (ωf - ωi) / t
where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken. The final angular velocity can be calculated by converting 3/4 of a turn into radians:
θ = (3/4) * 2π radians = (3/4) * 6.28 = 4.71 radians
Since the wheel comes to rest, the final angular velocity (ωf) is 0 rad/s. We're given the initial angular velocity (ωi) as 1.33 rad/s. Let's calculate the time (t) taken using the formula:
t = θ / ωi
t = 4.71 radians / 1.33 rad/s = 3.54 seconds
Now, we can substitute the values of ωf, ωi, and t into the equation for angular acceleration to find α:
α = (0 - 1.33 rad/s) / 3.54 s = -0.376 rad/s^2
Finally, the torque (τ) exerted on the wheel can be calculated using the equation:
τ = I * α
τ = 1.463 kg·m^2 * -0.376 rad/s^2 = -0.550 N·m
The average torque exerted on the wheel is -0.550 N·m. The negative sign indicates that the torque is in the opposite direction to the initial angular velocity.