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A uniform circular disc of radius 50cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0rads2. Its net acceleration in ms2 at the end of 2.0s is approximately:
A. 8.0
B. 7.0
C. 6.0
D. 3.0

Answer
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Hint: Use the kinematic equation relating final angular velocity, initial angular velocity, angular acceleration and time to determine the final angular velocity of the disc at the end of 2.0s. Determine the linear centripetal and tangential acceleration and then resultant of these two accelerations.

Formulae used:
The kinematic equation relating final angular velocity ω, initial angular velocity ω0, angular acceleration α and time t is
ω=ω0+αt …… (1)
The centripetal acceleration aC of an object is
aC=ω2R …… (2)
Here, ω is the angular velocity of the object and R is the radius of the circular path.
The tangential acceleration at of an object is
at=αR …… (3)
Here, α is the angular acceleration of the object and R is the radius of the circular path.

Complete step by step solution:
We have given that the radius of the disc is 50cm.
R=50cm
The angular acceleration of the disc is 2.0rads2.
α=2.0rads2
The initial angular velocity of the object is zero as it starts from rest.
ω0=0radss1
Let us determine the angular velocity of the disc at the end of 2.0s using equation (1).
Substitute 0radss1 for ω0, 2.0rads2 for α and 2.0s for t in equation (1).
ω=(0radss1)+(2.0rads2)(2.0s)
ω=4rads1

Hence, the angular velocity of the disc at the end of 2.0s is 4rads1.
Let us determine the centripetal acceleration of the disc.
Substitute 4rads1 for ω and 50cm for R in equation (2).
aC=(4rads1)2(50cm)
aC=(4rads1)2[(50cm)(102m1cm)]
aC=8m/s2
Let us determine the tangential acceleration of the disc.
Substitute 2.0rads2 for α and 50cm for Rin equation (3).
at=(2.0rads2)(50cm)
at=(2.0rads2)[(50cm)(102m1cm)]
at=1m/s2
The net linear acceleration of the disc is the resultant of the centripetal and tangential acceleration of the disc.
a=aC2+at2
Substitute 8m/s2 for aC and 1m/s2 for at in the above equation.
a=(8m/s2)2+(1m/s2)2
a=64+1
a=65
a8ms2

Therefore, the net acceleration of the disc is approximately 8ms2.Hence, the correct option is A.

Note: The students may think that we have given the angular acceleration of the object and we need to determine the angular acceleration at the end of 2 seconds. But the student should read the question carefully because we have asked the net acceleration in meter per second square which is the unit of linear acceleration. So, we have to determine the linear acceleration at the end of 2 seconds.
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