Q.A stone of mass $0.25\ \text{kg}$ tied to the end of a string is whirled round in a circle of radius $1.5\ \text{m}$ with a speed of $40\ \text{rev./min}$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200\ \text{N}$?

Concept and intuition:

To successfully go around the “globe of death,” the motorcyclist must not lose contact with the track at the highest point of the loop. At that point, the normal force can be zero at minimum — the centripetal force is then provided entirely by gravity. This gives a minimum speed at the top. Then, using conservation of mechanical energy from the start (at height $h$) to the top of the loop (height $2r$), we can find the necessary starting height. The ratio $h/r$ emerges naturally.

Step-by-step solution:

1. Minimum speed at the top of the loop At the top of the globe (radius $r$), the motorcyclist is upside-down. For the minimum safe speed, the normal force from the track is zero, so the only force providing the centripetal acceleration is gravity:

Cancelling $m$ gives:

2. Conservation of energy The motorcyclist starts from rest at height $h$ above the bottom of the incline. The bottom of the incline is at the same level as the bottom of the globe (we assume the track is smooth and frictionless). At the top of the globe, the height above the bottom is $2r$ (diameter). Energy conservation: $$$$

The key idea is that the particle X moves in a circle because the electrostatic force from Y provides the necessary centripetal force. Since the motion is uniform circular motion with a given frequency, we can relate the speed to the radius and frequency, then solve for mass.

Step-by-step reasoning:

1. Identify the force providing centripetal acceleration. Particle X (charge $+Q$) and particle Y (charge $-Q$) attract each other via Coulomb’s law. The magnitude of the electrostatic force is

This force acts as the centripetal force keeping X in a circle of radius $R$ around Y.

2. Write the centripetal force condition. For a particle of mass $m$ moving in a circle of radius $R$ with speed $v$, the required centripetal force is

Setting $F_e = F_c$ gives

Simplify by multiplying both sides by $R$:

3. Relate speed to frequency. The frequency $\nu$ (often denoted $f$ in other contexts) is the number of revolutions per second. In one revolution, the particle travels the circumference $2\pi R$, so the speed is

Substitute this into equation (1):

4. Solve for mass $m$. …

The key idea is that for the coin to rotate with the disc (without slipping), the static friction must provide the necessary centripetal force. Static friction can adjust up to a maximum value of $\mu N = \mu mg$. As the coin is placed farther from the centre, the required centripetal force $m\omega^2 r$ increases. The maximum safe radius occurs when these two forces are equal.

Let's work through it step by step:

1. Identify the forces in the rotating frame. In the inertial frame, the only horizontal force on the coin is static friction $f_s$, directed toward the centre. This friction provides the centripetal acceleration needed for circular motion:

where $m$ is the coin's mass, $\omega$ is the angular speed, and $r$ is the distance from the centre.

2. Recognize the limit of static friction. Static friction can vary from zero up to a maximum value:

Here the normal force $N$ equals the weight $mg$ (since the disc is horizontal), so

3. Set the condition for no slipping. For the coin to stay in place without sliding outward, the required centripetal force must not exceed the maximum available friction:

Cancel $m$ (it doesn't matter — a heavier coin needs more force but also has more friction):

4. Solve for the maximum radius. …

Concept & Intuition

When an object moves along a curved path, its total acceleration has two perpendicular components:

• Centripetal acceleration ($a_c$) points toward the center, responsible for changing direction. Its magnitude is $v^2/r$.
• Tangential acceleration ($a_t$) points along the tangent, responsible for changing speed.

The net acceleration is the vector sum of these two. If we know the angle between the net acceleration and the centripetal component, we can relate $a_c$ and $a_t$ using trigonometry. Here, the angle is $30^\circ$ and $a_t = 10\ \mathrm{m/s^2}$, so we can find $a_c$, then solve for $v$.

Step-by-step solution

1. Set up the vector relationship The centripetal acceleration $a_c$ is perpendicular to the tangential acceleration $a_t$. The net acceleration $\vec{a}_{\text{net}}$ is the diagonal of the rectangle formed by these two. The angle between $\vec{a}_{\text{net}}$ and $a_c$ is given as $30^\circ$. From the right triangle:

because $a_t$ is opposite the $30^\circ$ angle and $a_c$ is adjacent.

2. Solve for centripetal acceleration

3. Relate centripetal acceleration to speed For circular motion:

with $r = 5\ \mathrm{m}$. Substitute:

For uniform circular motion the centripetal acceleration is

The frequency is numerically equal to the radius, $f=r$, so

Substituting $r=\dfrac{5}{\pi}\ \text{m}$: …

The table is horizontal and smooth. The particle's weight $mg$ is balanced by the normal reaction $N$; both are vertical and do not contribute to the horizontal (centripetal) direction. …

Concept: Circular motion under an attractive inverse-square central force (identical in structure to gravitation/the Bohr atom).

The centripetal force has magnitude K/r² (the minus sign only means it points inwards):

m v²/r = K/r² → m v² = K/r.

Kinetic energy: KE = ½ m v² = K/(2r).

Potential energy: F = −dU/dr with F_r = −K/r² gives U(r) = −K/r (taking U(∞) = 0).

Total energy: E = KE + PE = K/(2r) − K/r = −K/(2r). …

Concept: car on a level circular road - the ONLY horizontal force available is static friction, and it must supply the centripetal force.

Required centripetal force = m v^2 / r, provided by friction f.

But static friction cannot exceed its limiting value: f <= mu_s N.

So m v^2 / r = f <= mu_s N (which also gives the maximum safe speed v_max = sqrt(mu_s g r), when the equality holds). …

Step 1 — The governing concept.

For a system of particles,

The forces the two particles exert on each other are internal — by Newton's third law they are equal and opposite and cancel in the sum. So $\vec{F}_{\text{ext}} = 0$, which means $\vec{a}_{cm} = 0$ and $\vec{v}_{cm}$ is constant.

Step 2 — Use the initial condition.

Both particles are initially at rest, so at $t=0$:

A constant that starts at zero is zero for all time:

Step 3 — Cross-check with the given speeds.

The speeds $v$ and $2v$ are not decoration — they are consistent with (and follow from) momentum conservation. Zero total momentum forces

i.e. the lighter particle moves faster, exactly as expected. Feeding this back: …

The key idea here is that angular momentum depends not just on how fast the particle is moving, but on the perpendicular distance from the reference point to the particle's line of motion. When a particle moves along a straight line with constant speed, that perpendicular distance doesn't change — so the angular momentum stays the same.

Let's see why this works step by step.

1. Define angular momentum for a particle in linear motion. For a particle of mass $m$ moving with velocity $\vec{v}$, the angular momentum about a point O is

where $\vec{r}$ is the position vector from O to the particle. The magnitude is

where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$.

2. Interpret $r \sin\theta$ geometrically.

   Look at the triangle formed by O, the particle's current position, and the line of motion. The quantity $r \sin\theta$ is exactly the perpendicular distance from O to the line along which the particle moves — call it $d$. This distance is constant for a fixed point O and a fixed straight line path.

   Tip

   Instead of tracking the changing $r$ and $\theta$ separately, just spot that $r \sin\theta$ is the lever arm — the shortest distance from O to the line of motion. That distance never changes as the particle slides along the line.

3. Apply to the given motion. …