Learn law of conservation of angular momentum. Practice to calculate conservation of angular momentum when external torque is zero and relation between moment of inertia and angular velocity.
Angular momentum of \(i^{th}\) particle is given by:
\(L_i=m_iv_ir_i\)
where, \(r_i\) is the perpendicular distance of \(i^{th}\) particle from the axis and \(v_i\) is the velocity of \(i^{th}\) particle.
If axis of rotation is same for all particles:
\(L_{body}=\Sigma L_i\)
\(=\Sigma m_iv_ir_i\)
\(=\Sigma m_i(r_i\omega)r_i\)
\(L_{body}=(\Sigma m_ir_i^2)\omega\)
\(L_{axis}=I_{axis}\,\omega\)
A \(2mr^2\omega\)
B \(3mr^2\omega\)
C \(\dfrac{1}{2}mr^2\omega\)
D \(4mr^2\omega\)
\(\vec{\tau}_{net}=0\)
\(\dfrac{dL}{dt}=0\)
then, \(I_1\omega_1=I_2\omega_2\)
\(K=\dfrac{1}{2}I\omega^2\)
\(\Rightarrow K=\dfrac{1}{2}(I\omega)\omega\) or \(\dfrac{1}{2} \dfrac{(I^2\omega^2)}{I} \)
\(\Rightarrow K=\dfrac{1}{2}L\omega\) or \(\dfrac{1}{2} \dfrac{L^2}{I} \)
A Its angular velocity gets doubled and kinetic energy remains same.
B Its angular velocity remains same and kinetic energy gets doubled.
C Its angular velocity and kinetic energy gets halved.
D Its angular velocity and kinetic energy gets doubled.
\(\vec{\tau}_{sys}=0\)
A \(\dfrac{mvR}{I}\)
B \(\dfrac{mvR}{MR^2+I}\)
C \(\dfrac{2mv}{MR}\)
D \(\dfrac{mv}{MR}\)
\(\Rightarrow \dfrac{d\;L_{axis}}{dt}=0\)
\(\Rightarrow L_{axis}\) = constant
A Double
B Remain same
C Reduce to half
D Become 4 times
here,
\(\vec{F_{12}}\) is a force applied on particle 1 due to particle 2
\(\vec{F_{21}}\) is a force applied on particle 2 due to particle 1
if \(\vec{F_{12}}=\vec{F}\)
then \(\vec{F_{21}}=-\vec{F}\)
Torque, \(\tau_0=\vec{r_1}×\vec{F_{12}}+\vec{r_2}×\vec{F_{21}}\)
\(=\vec{r_1}×\vec{F}+\vec{r_2}×(-\vec{F})\)
\(=\vec{r_1}×\vec{F}-\vec{r_2}×\vec{F}\)
\(=(\vec{r_1}-\vec{r_2})×\vec{F}\)
\(\because \) The vector \((\vec{r_1}-\vec{r_2})\) is along \(\vec{F}\)
\(\therefore\tau_0=0\)
Torque \(\vec{\tau}\) due to all pairs of internal forces about any point is zero.
\(\therefore\vec{\tau}_{int}=0\)
If external torque is zero,
\(\vec{\tau}=\dfrac{d\vec{L}}{dt}\)
\(\Rightarrow\sum \vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}\)
If \(\sum \vec {\tau}_{ext}=0\)
then \(L_{sys}\) is constant.
Thus, if net external torque on a system is zero, its angular momentum remains conserved.
A \(200\;rev/min\)
B \(100\;rev/min\)
C \(300\;rev/min\)
D \(150\;rev/min\)
\(\Sigma\vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}\)
then \(\vec{L}_{sys}\) is a constant.
A \(\dfrac{mv}{(m+M)L}\)
B \(\dfrac{2mv}{ML}\)
C \(\dfrac{2mv}{(M+m)L}\)
D \(\dfrac{6mv}{(3M+4m)L}\)
As \(\overrightarrow{\tau}_{ext}=0\)
\(L_{sys}\) remains conserved.
When the event is impulsive, then the forces become internal and do not produce any torque. Thus, angular momentum of a body does not change and remains conserved.
Mechanical energy of a system is conserved if there are no external forces and the work done by internal non-conservative forces, is zero.
Mechanical energy of a system is not conserved during impulsive situations like collision, jumping and landing.
The conservation of mechanical energy can be applied after the collision/event has taken place and dissipative and non-conservative forces are absent.
A \(\dfrac{\pi}{2}\)
B \(\dfrac{\pi}{6}\)
C \(\dfrac{\pi}{3}\)
D \(\dfrac{\pi}{4}\)