- #1
louise
- 3
- 0
A load of mass 250kg is required to be lifted by a means of a winding drum and cable. The mass will be initially at rest, accelerated uniformly upwards for 4 seconds and then decelerated uniformly for one second. such that a final height of 10.5 metres is gained.
The winding drum has a mass of 225kg, a diameter of 1.5m and a radius of gyration of 320mm. The bearings of the drum have a constant frictional torque of 5Nm.
Calculate:
a) The maximum velocity reached by the mass.
b) The maximum angular velocity of the winding drum.
c) The total work done whilst accelerating the load upwards.
d) The input torque to the driving motor whilst accelerating the load.
e) The average power required whilst accelerating the load.
f) The maximum power required from the drive motor.
My solutions so far:
a) Using: s=ut+1/2at^2 (1/2 being half and ^2 being squared)
a=2(s-ut)/t^
=2(10.5-0*4)/4^2
= 1.31m/s^2
Using: v=u+at
=0+1.13*4
=5.24m/s^2
b) Using: θ=((ω1+ω2)t)/2 UNSURE ON THIS ON FOR DEFINITE!
=((0+5.24)4)/2
=10.48radians?
θ/t=Angular Velocity
10.48/2∏ (∏ being used as Pi)
=16.46revs
c) NEED HELP I HAVE THIS EQUATION BUT I NEED TO FIGURE OUT ALL THE LETTERS FIRST
WD=mg(h2-h1)+1/2m(v^2-u^2)+1/2I(ω2^2-ω1^2)+(Fr*θ)
I know mK^2 =I m being the mass and K being the radius of gyration?
An to find h2 I use S=((u+v)t)/2
d)Ang Power =WD/t?
f)??
g) Max Power = T*ω1??
PROBABLY ALL WRONG - PLEASE HELP, QUITE BAFFLED AT THE MOMENT!
The winding drum has a mass of 225kg, a diameter of 1.5m and a radius of gyration of 320mm. The bearings of the drum have a constant frictional torque of 5Nm.
Calculate:
a) The maximum velocity reached by the mass.
b) The maximum angular velocity of the winding drum.
c) The total work done whilst accelerating the load upwards.
d) The input torque to the driving motor whilst accelerating the load.
e) The average power required whilst accelerating the load.
f) The maximum power required from the drive motor.
My solutions so far:
a) Using: s=ut+1/2at^2 (1/2 being half and ^2 being squared)
a=2(s-ut)/t^
=2(10.5-0*4)/4^2
= 1.31m/s^2
Using: v=u+at
=0+1.13*4
=5.24m/s^2
b) Using: θ=((ω1+ω2)t)/2 UNSURE ON THIS ON FOR DEFINITE!
=((0+5.24)4)/2
=10.48radians?
θ/t=Angular Velocity
10.48/2∏ (∏ being used as Pi)
=16.46revs
c) NEED HELP I HAVE THIS EQUATION BUT I NEED TO FIGURE OUT ALL THE LETTERS FIRST
WD=mg(h2-h1)+1/2m(v^2-u^2)+1/2I(ω2^2-ω1^2)+(Fr*θ)
I know mK^2 =I m being the mass and K being the radius of gyration?
An to find h2 I use S=((u+v)t)/2
d)Ang Power =WD/t?
f)??
g) Max Power = T*ω1??
PROBABLY ALL WRONG - PLEASE HELP, QUITE BAFFLED AT THE MOMENT!