- #1
r4vr4c
- 8
- 0
A flywheel is driven by a pulley system. friction in the system exerts a braking force of 2Nm. The moment of inertia is 400kgm2.
The forces in the pulley belt are 120N on the tight side and 80N on the slack side.
Pulley diameter is 0.2m
Calculate the acceleration of the wheel.
the pulley has the greater diameter but doesn't say the diameter of the motor but doesn't give a diameter for the motor. also doesn't say which way the flywheel is moving.
I have made several attempts using equations like the square root of I/m but i don't know the mass. i did 400/120 which i think gives me the radius of gyration. but as i already know the inertia i don't then thought i don't need this.
so i then did T= F x R
But as I'm not sure which is the side with the pulley i just guessed that it was the bigger force of 120N
120/0.1 = 1200Nm
T=I x a
a= T/I
1200/400 = 3rad/s2
or is the using the other 80N
or is it using the 80N and then subtracting the 1st from the second?
how does the 2N of braking force get accounted for?
The forces in the pulley belt are 120N on the tight side and 80N on the slack side.
Pulley diameter is 0.2m
Calculate the acceleration of the wheel.
the pulley has the greater diameter but doesn't say the diameter of the motor but doesn't give a diameter for the motor. also doesn't say which way the flywheel is moving.
I have made several attempts using equations like the square root of I/m but i don't know the mass. i did 400/120 which i think gives me the radius of gyration. but as i already know the inertia i don't then thought i don't need this.
so i then did T= F x R
But as I'm not sure which is the side with the pulley i just guessed that it was the bigger force of 120N
120/0.1 = 1200Nm
T=I x a
a= T/I
1200/400 = 3rad/s2
or is the using the other 80N
or is it using the 80N and then subtracting the 1st from the second?
how does the 2N of braking force get accounted for?