Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Two horizontal disks of mass \(M\) have the radii shown above. Disk A is attached to an axle of negligible mass spinning freely with angular velocity \(\omega_0\). Disk B, not attached to the axle and initially held at rest, is released and drops onto disk A. When both disks spin together without slipping, the angular velocity \(\omega_f\) of the disks is
Correct Answer: \(\boxed{\mathrm{D}}\)
Since there is no external torque acting on the two-disk system, angular momentum is conserved:
\(L_i=L_f\)
Initially, only Disk A rotates. Its radius is \(2R\), so its moment of inertia is
\(I_A=\frac{1}{2}M(2R)^2=2MR^2\)
Disk B has radius \(R\), so its moment of inertia is
\(I_B=\frac{1}{2}MR^2\)
Applying conservation of angular momentum,
\(I_A\omega_0=(I_A+I_B)\omega_f\)
Substituting the moments of inertia,
\(2MR^2\omega_0=\left(2MR^2+\frac{1}{2}MR^2\right)\omega_f\)
\(2MR^2\omega_0=\frac{5}{2}MR^2\omega_f\)
Solving for the final angular velocity,
\(\omega_f=\frac{2}{5/2}\omega_0=\frac{4}{5}\omega_0\)
Although mechanical energy decreases because friction brings the disks to a common angular speed, angular momentum remains conserved.
Therefore, the correct answer is (D).
What is the fewest number of the following conditions required to ensure that angular momentum is conserved?
I. Conservation of linear momentum
II. Zero net external force
III. Zero net external torque
Correct Answer: \(\boxed{\mathrm{B}}\)
Angular momentum is conserved whenever the net external torque on a system is zero:
\(\sum \tau_{\mathrm{ext}}=\frac{dL}{dt}\)
Therefore, if
\(\sum \tau_{\mathrm{ext}}=0\),
then
\(L=\text{constant}\)
A system may have zero net external force (or conserved linear momentum) and still experience a nonzero external torque. Thus, conditions I and II alone do not guarantee conservation of angular momentum.
The only condition required is zero net external torque.
Therefore, the correct answer is (B).
A bullet is moving with a velocity \(v_0\) when it collides with and becomes embedded in a wooden bar that is hinged at one end, as shown above. Consider the bullet and the wooden bar to be the system.
For this scenario, which of the following is true?
Correct Answer: \(\boxed{\mathrm{B}}\)
When the bullet becomes embedded in the wooden bar, the collision is perfectly inelastic. Therefore, kinetic energy is not conserved.
The hinge exerts a large external force on the bar during the collision, so the system’s linear momentum is not conserved:
\(\sum \vec{F}_{\mathrm{ext}}\neq0\)
However, taking moments about the hinge, the hinge force produces zero torque because its line of action passes through the pivot:
\(\tau_{\mathrm{hinge}}=rF\sin\theta=0\)
During the short collision, the net external torque about the hinge is negligible. Therefore,
\(\sum\tau_{\mathrm{ext}}=0\)
which implies
\(L_{\mathrm{initial}}=L_{\mathrm{final}}\)
Thus, angular momentum about the hinge is conserved, while linear momentum and kinetic energy are not.
Therefore, the correct answer is (B).
