Answer:
a) 3.265 kg-m^2
b) - 4.96 rad/s^2
c) 16.1944 N-m
d) 35.625
e) - 1.488 m/s^2
f) 67.1175 m.
Step-by-step explanation:
Given data:
Diameter of grindstone ( D ) = 0.650 m , Radius ( R ) = 0.325 m
mass of grindstone ( M ) = 55 kg
Radius of shaft ( r ) = 0.300 m
mass of shaft ( m ) = 4 kg
Initial Angular velocity = 450 rev/min = f = 7.5 rev/s = w =15π rad/s
time ( t ) = 9.50 secs
a) Determine the moment of inertia of the grindstone including the shaft
moment of inertia of grindstone = MR^2 / 2 = 55* (0.325)^2 / 2 = 2.905
moment of inertia of shaft = mr^2 = 4 *0.3^2 = 0.36
∴ moment of inertia including shaft = 2.905 + 0.36 = 3.265 kg-m^2
b) Determine the angular acceleration of the grindstone
∝ = - 15π / 9.5 ( i.e. angular velocity / time )
= - 4.96 rad/s^2 ( deceleration value )
c) Determine average torque done by friction to bring the grindstone to rest
Torque ( I * ∝ ) = 3.265 x 4.96 = 16.1944 N-m (magnitude)
d) Determine the number of revolutions before grindstone comes to rest
Total revolutions N before grindstone comes to rest
= ( f1 + f2)* t /2 = 7.5 * 9.5 / 2 = 35.625
Note : f2 = 0 as it comes to rest
f1 = 7.5 rev/s
e) Determine the Linear acceleration of the shaft in this process
This can be calculated using this relation
r * ∝ = 0.3 x (- 4.96 ) = - 1.488 m/s^2
f) Determine the linear distance travelled by the shaft
This can be calculated with the relation below
r * 2 * π * N = 0.3 * 2π * 35.625 = 67.1175 m.