# Angular momentum of a rolling cylinder about the point of contact

A cylinder of mass 5 kg and radius 10 cm is moving on a horizontal surface with velocity of centre of mass 5 m/s towards right and angular speed 10 rad/s (clockwise) . Find the angular momentum of the cylinder about the point of contact.

My approach: The angular velocity of the body about the point of contact will be 10 rad/s and the moment of inertia about the point will be 1.5 MR^2 . So angular momentum should be (moment of inertia)*(angular velocity)= 10 * 1.5 (5)(0.1)(0.1) giving 0.75 but the answer is 2.75 . I don't know where I have gone wrong.

You forgot to include angular momentum of center of mass.

Thus add $MR^2\omega$ to your answer

By the way angular momentum is not $I\omega$, it is $\vec{l_{com}}+I_{com}\vec{\omega}$

It is $I_{contact}\omega$ only for cases when point of contact is the point of instantaneous or fixed axis of rotation.

• Why will I add angular momentum of centre of mass ? And how come angular momentum of centre of mass is MR^2w . I think centre of mass is just a point it does not have any mass . A mass can be kept at centre of mass but centre of mass has no mass . Apr 16 '16 at 17:47

There are a few things in your question that do not add up, a cylinder of radius $R = 0.1 \mathrm m$ rolling, I assume without slipping, at a velocity $v_{\mathrm C}= 5 \mathrm{ms^{-1}}$ has an angular velocity $$\omega = \frac{v_{\mathrm C}}{R} = \frac{5}{0.1}\mathrm{\frac{rad}{s}} = 50\mathrm{\frac{rad}{s}} \neq 10\mathrm{\frac{rad}{s}}$$

Let's assume that the velocity is correct and $\omega = 50\mathrm{\frac{rad}{s}}$. We know that the moment of inertia about the center $\mathrm C$ is: $$I_\mathrm{C}=\int_M r^2 \mathrm dm = \int_0^R r^2 \left(\frac{M}{R^2}((r+\mathrm dr)^2 - r^2)\right) = \frac{2M}{R^2}\int_0^Rr^3\mathrm dr=\frac{1}{2}MR^2$$ The Parallel Axis Theorem states that the moment of inertia about a parallel axis passing through $\mathrm P$ (point of contact) would be: $$I_{\mathrm P} = I_{\mathrm C} + Md^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$$ This makes the angular momentum $L_{\mathrm P} = I_{\mathrm P}\omega =\frac{3}{2}MR^2\omega = 3.75$.

• It isn't the answer that OP was expecting but, after the angular velocity, his numbers lost some credibility. May 20 '18 at 11:01