The centre of gravity of a body on the earth coincides with its centre of mass for a small object whereas for an extended object it may not. What is the qualitative meaning of small and extended in this regard? For which of the following two coincides? A building, a pond, a lake, a mountain?
When the vertical height of the object is very small as compared to the earth's radius, we call the object small, otherwise it is extended.
(i) Building and pond are small objects.
(ii) A deep lake and a mountain are examples of extended objects.
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)