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series111
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Homework Statement
a light shaft carries a steel disk 400m dia, 50mm thick,density 7800kg/m_3 calculate :
(a) if the frictional resistance is equivalent to a torque of 1.5Nm, determine the braking torque to bring the disk to rest from 43.98 rad/s in 4 secs.
(b) what would be the time for the system to stop, if the motor was turned off with no braking applied ?
i also had work out the change in angular momentum which = 148 kgm_2/s
and change in angular kinetic energy = 3.7 KJ
Homework Equations
acceleration = final velocity - initial velocity/ time
Torque = moment of inertia x acceleration
The Attempt at a Solution
Question (a)
acceleration = 0 - 43.98/4 = 10.99 rad/s
Torque = moment of inertia x acceleration
: moment of inertia = torque / acceleration = 1.5Nm / 10.99 rad/s = 136.48 x 10 ^-3
Torque = moment of inertia x acceleration = 136.48 x 10 ^-3 x 10.99 rad/s = 1499.91 Nm
Question (b)
acceleration = final velocity - initial velocity/ time
: time = final velocity - initial velocity/acceleration = 0 - 43.98/10.99 = 4 sec
this does not look right to me but where have i went wrong any help would be most appreciated thanks again...
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