Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?
The moment of inertia of a body is given by $$I=\Sigma m_i r_i^2$$ [sum of moment of inertia of each constituent particles]
All the mass in a cylinder lies at distance $$R$$ from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than $$R$$.
The variation of angular position $$\theta$$, of a point on a rotating rigid body, with time $$t$$ is shown in figure. Is the body rotating clockwise or anti-clockwise?
As the slope of $$\theta$$-t graph is positive and positive slope indicates anti-clockwise rotation which is traditionally taken as positive.
A uniform cube of mass $$m$$ and side $$a$$ is placed on a frictionless horizontal surface. A vertical force $$F$$ is applied to the edge as shown in figure. Match the following (most appropriate choice)
| (a) | $$ m g / 4<F<m g / 2 $$ |
(i) | Cube will move up. |
|---|---|---|---|
| (b) | $$F > mg/2$$ | (ii) | Cube will not exhibit motion. |
| (c) | $$F > mg$$ | (iii) | Cube will begin to rotate and slip at A. |
| (d) | $$F=mg/4$$ | (iv) | Normal reaction effectively at a/3 from A, no motion. |
Consider the below diagram
Moment of the force $$F$$ about point $$A, \tau_1=F \times a$$ (anti-clockwise)
Moment of weight $$m g$$ of the cube about point $$A$$,
$$\tau_2=m g \times \frac{a}{2}(\text { clockwise })$$
Cube will not exhibit motion, if $$\tau_1=\tau_2$$
($$\because$$ In this case, both the torque will cancel the effect of each other)
$$\therefore \quad F \times a=m g \times \frac{a}{2} \Rightarrow F=\frac{m g}{2}$$
$$\begin{aligned} &\text { Cube will rotate only when, } \tau_1>\tau_2\\ &\Rightarrow \quad F \times a>m g \times \frac{a}{2} \Rightarrow F>\frac{m g}{2} \end{aligned}$$
$$\begin{aligned} & \text { Let normal reaction is acting at } \frac{a}{3} \text { from point } A \text {, then } \\ & \qquad m g \times \frac{a}{3}=F \times a \text { or } F=\frac{m g}{3} \quad \text{For no motion} \end{aligned}$$
$$\begin{aligned} &\text { When } F=\frac{m g}{4} \text { which is less than } \frac{m g}{3} \text {,. }\quad\left(F<\frac{m g}{3}\right) \end{aligned}$$
there will be no motion.
$$\therefore \quad$$ (a) $$\rightarrow$$ (ii) (b) $$\rightarrow$$ (iii) (c) $$\rightarrow$$ (i) (d) $$\rightarrow$$ (iv)
A uniform sphere of mass $$m$$ and radius $$R$$ is placed on a rough horizontal surface (figure). The sphere is struck horizontally at a height $$h$$ from the floor. Match the following
| (a) | $$h=R/2$$ | (i) | Sphere rolls without slipping with a constant veloci8ty and no loss of energy. |
|---|---|---|---|
| (b) | $$h > R$$ | (ii) | Sphere spins clockwise, loses energy by friction. |
| (c) | $$h > 3R/2$$ | (iii) | Sphere spins anti-clockwise, loses energy by friction. |
| (d) | $$h=7R/5$$ | (iv) | Sphere has only a translational motion, loses energy by friction. |
Consider the diagram where a sphere of $$m$$ and radius $$R$$, struck horizontally at height $$h$$ above the floor
The sphere will roll without slipping when $$\omega=\frac{v}{r}$$, where, $$v$$ is linear velocity and $$\omega$$ is angular velocity of the sphere.
Now, angular momentum of sphere, about centre of mass
[We are applying conservation of angular momentum just before and after struck]
$$\begin{aligned} m v(h-R) & =I \omega=\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right) \\ \Rightarrow \quad m v(h-R) & =\frac{2}{5} m v R \\ h-R & =\frac{2}{5} R \Rightarrow h=\frac{7}{5} R \end{aligned}$$
Therefore, the sphere will roll without slipping with a constant velocity and hence, no loss of energy, so (d) $$\rightarrow$$ (i)
Torque due to applied force, $$F$$ about centre of mass
$$\tau=F(h-R)\quad \text{(clockwise)}$$
For $$\tau=0, h=R$$, sphere will have only translational motion. It would lose energy by friction.
Hence, (b) $$\rightarrow$$ (iv)
The sphere will spin clockwise when $$\tau>0 \Rightarrow h>R$$
Therefore, (c) $$\rightarrow$$ (ii)
The sphere will spin anti-clockwise when $$\tau<0 \Rightarrow h< R$$, (a) $$\rightarrow$$ (iii)
The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?
No, not necessarily.
Given, $$\quad \sum_\limits i F_i \neq 0$$
$$\text { The sum of torques about a certain point } O, \sum_\limits i \mathbf{r}_i \times \mathbf{F}_i=0$$
$$\begin{aligned} &\text { The sum of torques about any other point } \mathrm{O}^{\prime}\\ &\sum_i\left(\mathbf{r}_i-\mathbf{a}\right) \times \mathbf{F}_i=\sum_i \mathrm{r}_i \times \mathbf{F}_i-\mathbf{a} \times \sum_i \mathrm{~F}_i \end{aligned}$$
Here, the second term need not vanish.
Therefore, sum of all the torques about any arbitrary point need not be zero necessarily.