Engineering Mechanics - PKRB: Impulse and Momentum - Discussion
Discussion Forum : PKRB: Impulse and Momentum - General Questions (Q.No. 11)
11.
A cord of negligible mass is wrapped around the outer surface of the 2-kg disk. If the disk is released from rest, determine its angular velocity in 3 s.
= 183.9 rad/s = 735 rad/s = 245 rad/s = 263 rad/sDiscussion:
2 comments Page 1 of 1.
KS37 said:
7 years ago
Mg-T=Ma --> eqn1
a=r*c --> eqn 2.
Where c= angular acceleration.
Torque about centre of disc = T*R = I*c
I=moment of inertia.
Eqn 2 is from assumption no slip at the point of contact.
From 1 and 2 and 3 we get C=2g/3r and angular velocity = C * t =2 * 9.81/0.08 = 245.25.
a=r*c --> eqn 2.
Where c= angular acceleration.
Torque about centre of disc = T*R = I*c
I=moment of inertia.
Eqn 2 is from assumption no slip at the point of contact.
From 1 and 2 and 3 we get C=2g/3r and angular velocity = C * t =2 * 9.81/0.08 = 245.25.
(1)
KS37 said:
7 years ago
Mg-T=Ma --> eqn1
a=r*c --> eqn 2.
Where c= angular acceleration.
Torque about centre of disc = T*R = I*c
I=moment of inertia.
Eqn 2 is from assumption no slip at the point of contact.
From 1 and 2 and 3 we get C=2g/3r and angular velocity = C * t =2 * 9.81/0.08 = 245.25.
a=r*c --> eqn 2.
Where c= angular acceleration.
Torque about centre of disc = T*R = I*c
I=moment of inertia.
Eqn 2 is from assumption no slip at the point of contact.
From 1 and 2 and 3 we get C=2g/3r and angular velocity = C * t =2 * 9.81/0.08 = 245.25.
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