Circular Motion — JEE Main practice

15 questions

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Sample questions with solutions

Q1 · 2026

Two cars A and B each of mass $10^3$ kg are moving on parallel tracks separated by a distance of $10$ m, in same direction with speeds $72$ km/h and $36$ km/h. The magnitude of angular momentum of car A with respect to car B is ________ J·s.

A.
$3.6\times10^5$
B.
$10^5$
C.
$3\times10^5$
D.
$2\times10^5$

Answer: B

Convert both speeds into SI units.
$v_A=72\times\frac{5}{18}=20\ \mathrm{m/s}$ $v_B=36\times\frac{5}{18}=10\ \mathrm{m/s}$
Since both cars move in the same direction, the relative velocity of car A with respect to car B is
$\vec{v}_{AB}=\vec{v}_A-\vec{v}_B$
Therefore,
$v_{AB}=20-10=10\ \mathrm{m/s}$
The angular momentum of a particle about a point is given by
$\vec{L}=\vec{r}\times\vec{p}$
where $\vec{p}=m\vec{v}$ .
Since the cars move on parallel tracks, the perpendicular distance between their lines of motion is
$r_\perp=10\ \mathrm{m}$
Hence,
$L=m.v_{\text{relative}}.r_\perp$
Substitute the given values.
$L=(10^3)\times10\times10$ $L=10^5\ \mathrm{J\cdot s}$
Therefore, the magnitude of the angular momentum of car A with respect to car B is
$10^5\ \mathrm{J\cdot s}.$
Hence, the correct option is B.

Q2 · 2026

A large drum having radius $R$ is spinning around its axis with angular velocity $\omega$ , as shown in figure. The minimum value of $\omega$ so that a body of mass $M$ remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass $M$ as $\mu$ , is:

A.
$\sqrt{\dfrac{2g}{\mu R}}$
B.
$\sqrt{\dfrac{g}{2\mu R}}$
C.
$\sqrt{\dfrac{g}{\mu R}}$
D.
$\sqrt{\dfrac{\mu g}{R}}$

Answer: C

When the drum rotates, three forces act on the body:
- Weight, $\mathrm{Mg}$ , acts vertically downward.
- Normal reaction, $N$ , acts horizontally toward the centre of the drum.
- Friction, $f$ , acts upward to prevent the body from sliding downward.
The normal reaction provides the required centripetal force.
$N=M R\omega^2$
For the body to remain at rest relative to the rotating drum, the upward friction must balance its weight.
$f=\mathrm{Mg}$
Static friction can adjust only up to its maximum value.
$f_{\max}=\mu N$
Therefore,
$\mathrm{Mg}\leq\mu N$
Substitute the expression for the normal force.
$\mathrm{Mg}\leq\mu(MR\omega^2)$
Cancelling $M$ ,
$g\leq\mu R\omega^2$
Rearranging,
$\omega^2\geq\frac{g}{\mu R}$
Hence, the minimum angular speed is
$\omega_{\min}=\sqrt{\frac{g}{\mu R}}.$
Therefore, the correct option is C.

Q3 · 2026

In case of vertical circular motion of a particle by a thread of length $𝒓$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point $(A)$ of the circular path is $(g =$ gravitational acceleration $)$

A.
$\sqrt{4gr}$
B.
$\sqrt{\dfrac{5}{2}gr}$
C.
$\sqrt{\dfrac{7}{2}gr}$
D.
$\sqrt{5gr}$

Answer: C

Let the point where the tension becomes zero be P.

At point P, the forces acting along the radius are:

Tension, (T), directed towards the centre.
The radial component of weight, (\mathrm{mg}\sin30^\circ).

Apply the centripetal force equation.

T+\mathrm{mg}\sin30^\circ = \frac{\mathrm{m}v_P^2}{r}.

Since the tension is zero,

\begin{aligned} \mathrm{mg}\left(\frac{1}{2}\right) &= \frac{\mathrm{m}v_P^2}{r}, \\ v_P^2 &= \frac{gr}{2}. \end{aligned}

[ \qquad\cdots(1) ]

Find the height of point P above the bottom point A.

The centre of the circle is at a height (r) above A.

Point P is an additional height of

\begin{aligned} r\sin30^\circ &= \frac{r}{2}. \end{aligned}

Hence,

\begin{aligned} h &= r+\frac{r}{2} \\ &= \frac{3r}{2}. \end{aligned}

Apply the law of conservation of mechanical energy between the bottom point A and point P.

Taking the potential energy at A as zero,

\frac{1}{2}\mathrm{m}v_A^2 = \frac{1}{2}\mathrm{m}v_P^2 + \mathrm{m}gh.

Substitute the values of (v_P^2) and (h).

\begin{aligned} \frac{v_A^2}{2} &= \frac{1}{2}\left(\frac{gr}{2}\right) + g\left(\frac{3r}{2}\right) \\ &= \frac{gr}{4} + \frac{3gr}{2}. \end{aligned}

\begin{aligned} v_A^2 &= \frac{gr}{2} + 3gr \\ &= \frac{7gr}{2}. \end{aligned}

Therefore,

\begin{aligned} v_A &= \sqrt{\frac{7}{2}gr}. \end{aligned}

Hence, the correct option is C.

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