- #1
donniemateno
- 45
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im struggling abit with this question. i have a formula but it doesn't really go with my question
a brake disc has the following specification
mean radius 0.16m
force applied to each pad 5045N
U = 0.35
brake disc rotational speed is 550rev/min
work out its frictional torque
work done a min by this torque
heat energy generated pers second
i science for motor vehicle engineers by peter twigg and the book has a very similar example. but it has a brake cycliner in the equation which i dne have and hydraulic oil pressure.
i have the formula total braking force per disc = n x U x P x A (n)
in my case i take that to be 550 x 0.35 x 5.045Kn x 0.16
this gives me 155N of frictional torque per disc.
am i going along the right sort of lines or am i using the completely wrong forumla?
looking around a lot fo other examples have time bt as i dnt have this I am abit lost.
any help would be very much appriated :)
a brake disc has the following specification
mean radius 0.16m
force applied to each pad 5045N
U = 0.35
brake disc rotational speed is 550rev/min
work out its frictional torque
work done a min by this torque
heat energy generated pers second
i science for motor vehicle engineers by peter twigg and the book has a very similar example. but it has a brake cycliner in the equation which i dne have and hydraulic oil pressure.
i have the formula total braking force per disc = n x U x P x A (n)
in my case i take that to be 550 x 0.35 x 5.045Kn x 0.16
this gives me 155N of frictional torque per disc.
am i going along the right sort of lines or am i using the completely wrong forumla?
looking around a lot fo other examples have time bt as i dnt have this I am abit lost.
any help would be very much appriated :)