A hemispherical shell of mass $m$ and radius $R$ is hinged at point $O$ and placed on a horizontal surface $M N$ as shown in the figure. A ball of mass $m$ moving with velocity $u$ inclined at an angle $\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ strikes the shell at point $A$ (as shown in the figure) and stops. What is the minimum speed $u$, if the given shell is to reach the horizontal surface $O P$?

enter image description here

The answer given is:

For no velocity will the hemisphere reach the top

The reason given is:

Initial angular momentum about 'O' is zero and also, there is no torque about 'O'. So, $\omega=0$

My doubt is, will there be a impulse from the ground on the hemisphere which can provide impulsive torque? How do we prove that there will be no impulse due to the ball hitting the shell?

  • 1
    $\begingroup$ Is OP just floating in the air? $\endgroup$ Commented Jun 14, 2020 at 17:02
  • 1
    $\begingroup$ Consider OP as fixed in air $\endgroup$ Commented Jun 14, 2020 at 17:05
  • 1
    $\begingroup$ Is the collision of the ball with the shell elastic or inelastic? $\endgroup$
    – Puk
    Commented Jun 14, 2020 at 17:15
  • $\begingroup$ @Puk I don't think collision really matters as there is no torque. $\endgroup$ Commented Jun 14, 2020 at 17:33
  • $\begingroup$ Whether or not there is torque depends on whether or not the collision is elastic. $\endgroup$
    – Puk
    Commented Jun 14, 2020 at 17:34

3 Answers 3


From my understanding of rotational mechanics, the net torque non-zero if:

  1. The force applied on the body is non zero.
  2. Torque $\vec{\tau}=r \times \vec{F}$, with its magnitude as $rF\sin{\theta}$, so, for the torque to be non-zero, the force vector applied should not be parallel or antiparallel to the displacement vector from the hinge point (from $\sin{\theta}\neq{0}$).
  3. The force is applied at some distance from the hinge, and not right at the hinge (from r not equal to zero).

In your example, the second point is not fulfilled. The hinge point lies directly in the line of impulse.

So, the net torque on the shell due to applied impulse will be zero.

That is also why the initial angular momentum about the hinge point is said to be zero. The initial angular momentum (due to the point) equals $mvr$, where $m$ is the mass of the particle, $v$ is the initial velocity, and $r$ is the perpendicular distance between the velocity line of the particle, and the hinge point.

In this case, since the velocity line passes directly through the hinge point, $r=0$, and initial angular momentum will also be zero.

To get some amount of impulsive torque, the value of theta will have to be increased. (Here, I'm assuming that the shell can only rotate in the clockwise direction because of the fixed plank MN) If we do that, a net clockwise impulsive torque will be generated, and the shell will start to rotate.

  • $\begingroup$ I've no idea how to use MathJax yet, though, so theta, and X are used, instead of their correct symbols. I'll be grateful for any edits! $\endgroup$
    – VVidyan
    Commented Jun 15, 2020 at 7:26
  • 2
    $\begingroup$ Hi there! On math.meta.stackexchange.com/questions/5020/… you can find the basics you need for Mathjax. There is a reference included to a more extensive use. $\endgroup$ Commented Jun 15, 2020 at 8:32

Since torque about hinge point $O$ is zero due to zero moment arm so impulse will not be able to make any force couple and it will be balanced by net opposite force by rod $OP$ thus maintaining translational equilibrium.

Regarding your question there will be impulse exerted by ball but it will not cause any rotation and will be balanced by opposite force from rod to maintain translational equilibrium.


Let's assume that OP is a piece of the Earth. Obviously, a force is exerted on the semi-circle (let's assume the semi-circle is rigid but deformable): $F=\frac{\Delta p}{\Delta t}$, in the direction AO (though for a short while). $\Delta p$ is the change in momentum of the ball. It's only the vertical component of this force that provides the force for the torque (a torque that's also there for a short while, but enough to give the semi-circle a rotational velocity which is diminished by gravity). The horizontal component isn't involved in the movement of the semi-circle. The horizontal component of the force of impact on the semi-circle is counteracted by the reaction force of the semi-circle to the left.

The vertical component of the force is $F_v=\frac{1}{{\sqrt{5}}}\frac{\Delta p}{\Delta t}$. So there is a force to provide for a torque, which means the half-shell will start to rotate.

We don't have to calculate the torque explicitly though.
The semi-circle will have to rotate just $\frac{1}{2}\pi$ radians. The center of mass of the semi-circle lies $R(1-\frac{4}{3\pi})$ above the contact point of the semi-circle. This means the center of mass of the semi-circle will raise with an amount of $R+\frac{4}{3\pi}$ after a rotation of $\frac{1}{2}\pi$. So the potential energy of the semi-circle will have risen with $mgR(1+\frac{4}{3\pi})$. That's also the amount of kinetic energy, associated with the vertical component of the ball's velocity. The vertical velocity of the ball is $u_v=\frac{1}{\sqrt{5}}u$. So the ball's kinetic energy for the vertical velocity is $\frac{1}{10} m u^2$. This has to be equal to the difference in potential energy which we calculated to be $mgR(1+\frac{4}{3\pi})$:

$$\frac{1}{10} m u^2=mgR(1+\frac{4}{3\pi}),$$ so $$u=\sqrt{\frac{gR(1+\frac{4}{3\pi})}{10}}$$

So the vertical component of the ball makes the semi-circle move up and the horizontal component doesn't give rise to a torque. The energy and momentum associated with the horizontal component of the ball's velocity are, via the semi-circle, given to the Earth (MN).

If the semi-circle is Born-rigid (a physical impossibility) we have to write for the force: $F=\frac{dp}{dt}$, using infinitesimal differences in $p$ and $t$. Clearly, the force becomes infinite in this case, which is impossible.

  • $\begingroup$ If the collision is completely inelastic (i.e. the ball sticks to the shell), the net impulse applied to the shell at the point of impact is along $AO$, so there will be no net torque on the shell. In other words, there IS a torque associated with the tangential component of the force that negates the torque due to the normal component. $\endgroup$
    – Puk
    Commented Jun 14, 2020 at 20:08
  • $\begingroup$ @Puk The horizontal component of the force of impact on the semi-circle is canceled by the opposite force of the floor (MN) pushing back on the point of impact. So we're left with a vertical component only. Without the floor MN, no torque would be present. $\endgroup$ Commented Jun 14, 2020 at 21:00
  • $\begingroup$ No. There is no reason for the floor to apply any more force on the shell than the weight of the shell if the collision imparts no net torque on the shell to begin with. $\endgroup$
    – Puk
    Commented Jun 14, 2020 at 21:04
  • $\begingroup$ The horizontal component of the force pushes the sc to the right with the result that the floor pushes back on the sc with the same force so the horizontal component of the force of impact is canceled (if there is enough friction). The horizontal force is applied to the Earth (OP), so the Earth gains a little extra (rotational momentum. That's why I wrote that there is no torque without an underground.. $\endgroup$ Commented Jun 14, 2020 at 21:19
  • $\begingroup$ Nothing makes the shell rotate, because the net impulse imparted by the collision is in the direction of the pivot. Consider what would happen if there were no gravity or floor. Why would the shell move in the first place? $\endgroup$
    – Puk
    Commented Jun 14, 2020 at 21:23

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