Uniform Circular Motion MCQ Questions & Answers in Basic Physics | Physics

Since the body presses the surface with a force $$N$$ hence according to Newton's third law the surface presses the body with a force $$N.$$ The other force acting on the body is its weight $$mg.$$

For circular motion to take place, a centripetal force is required which is provided by $$\left( {mg + N} \right).$$
$$\therefore mg + N = \frac{{m{v^2}}}{r}$$
where $$r$$ is the radius of curvature at the top.
If the surface is smooth then on applying conservation of mechanical energy, the velocity of the body is always same at the top most point. Hence, $$N$$ and $$r$$ have inverse relationship. From the figure it is clear that $$r$$ is minimum for first figure, therefore $$N$$ will be maximum.
Note : If we do not assume the surface to be smooth, we cannot reach to a conclusion.

$$\eqalign{ & \tan \theta = \frac{{{v^2}}}{{rg}} \cr & \tan {12^ \circ } = \frac{{{{\left( {150} \right)}^2}}}{r} \times 10 \cr & \Rightarrow r = \frac{{2250}}{{0.2125}} = 10.6\,km \cr} $$

Let $$\omega $$ be the angular frequency of the system. The maximum acceleration of the system,
$$a = {\omega ^2}A = \left( {\frac{k}{{2m}}} \right)A\,\,\,\,\left[ {\omega = \sqrt {\frac{k}{{m + m}}} = \sqrt {\frac{k}{{2m}}} } \right]$$
The force of friction provides this acceleration.
$$\therefore f = ma = m\left( {\frac{{kA}}{{2m}}} \right) = \frac{{kA}}{2}$$

Centripetal acceleration $${a_n} = \frac{{{v^2}}}{r} = \frac{{{{10}^2}}}{2} = 50\,m/{s^2},$$      along negative $$x$$-axis.
$${\text{so}}\,\,\vec a = - 50\,\hat i - 1.5\,\hat j\,m/{s^2}$$

Tension at the highest point
$${T_{{\text{top}}}} = \frac{{m{v^2}}}{r} - mg = 2mg\left( {\therefore {v_{{\text{top}}}} = \sqrt {3gr} } \right)$$
Tension at the lowest point
$$\eqalign{ & {T_{{\text{bottom}}}} = 2mg + 6mg = 8mg \cr & \therefore \frac{{{T_{{\text{top}}}}}}{{{T_{{\text{bottom}}}}}} = \frac{{2mg}}{{8mg}} = \frac{1}{4}. \cr} $$

The velocity should be such that the centripetal acceleration is equal to the acceleration due to gravity $$\frac{{{v^2}}}{R} = g$$
$${\text{or}}\,v = \sqrt {gR} .$$

$$\eqalign{ & {\text{Given,}}\,\,\theta = {45^ \circ },\,r = 0.4\,m,\,g = 10\,m/{s^2} \cr & T\sin \theta = \frac{{m{v^2}}}{r}\,......\left( {\text{i}} \right) \cr & T\cos \theta = mg\,......\left( {{\text{ii}}} \right) \cr} $$
From equation (i) & (ii) we have,
$$\eqalign{ & \tan \theta = \frac{{{v^2}}}{{rg}} \cr & {v^2} = rg\quad \cr & \because \theta = {45^ \circ } \cr} $$
Hence, speed of the pendulum in its circular path,
$$v = \sqrt {rg} = \sqrt {0.4 \times 10} = 2\,m/s$$

Here, the horizontal component of tension provides the necessary centripetal force.
$$\therefore T\sin \theta = mr{\omega ^2}$$
From (i) and (ii)
$$\eqalign{ & T \times \frac{r}{L} = mr{\omega ^2}\,\,\,\,\,\left[ {\because \sin \theta = \frac{r}{L}} \right] \cr & \therefore \omega = \sqrt {\frac{T}{{mL}}} = \sqrt {\frac{{324}}{{0.5 \times 0.5}}} \cr & = \frac{{18}}{{0.5}} = 36\,rad/s \cr} $$

Force experienced by the particle, $$F = m{\omega ^2}R$$
$$\therefore \frac{{{F_1}}}{{{F_2}}} = \frac{{{R_1}}}{{{R_2}}}$$

For block $$A$$ to move in SHM.

$$mg - N = m{\omega ^2}x$$
where $$x$$ is the distance from mean position
For block to leave contact $$N = 0$$
$$ \Rightarrow mg = m{\omega ^2}x \Rightarrow x = \frac{g}{{{\omega ^2}}}$$