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Rotational Dynamics formulas

19 formulas and 19 common traps for MHT-CET Physics Rotational Dynamics, grouped by subtopic.

Full notes with worked examples

Kinematics of Circular Motion

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Angular Velocity, Linear Speed and Centripetal Acceleration

Circular motion

ω=2πn,v=ωr,ac=ω2r=v2r\omega = 2\pi n, \qquad v = \omega r, \qquad a_c = \omega^2 r = \frac{v^2}{r}

Constant Angular Acceleration

Angle in the nth second, from rest

θn=α2(2n−1)\theta_n = \frac{\alpha}{2}(2n - 1)

Tangential and Centripetal Acceleration Together

Net acceleration on a circle

a=(rα)2+(ω2r)2a = \sqrt{(r\alpha)^2 + (\omega^2 r)^2}

Common traps

Letting the masses matter

Same period means same ω; speed ratio is the radius ratio. The masses in the stem are there to be picked by mistake, and 'm₁ : m₂' is always an option.

Angle in the nth second versus angle in n seconds

'In the 3rd second' is one second's worth, α2(2n−1)\frac{\alpha}{2}(2n - 1); 'in 3 seconds' is 12α(9)\frac{1}{2}\alpha(9). Both land on options.

Using only the centripetal part for the net force

When the speed changes, the net force needs both components. 5 kg at at=4a_t = 4, ac=20a_c = 20 m/s² feels 5416=20265\sqrt{416} = 20\sqrt{26} N, not 100 N.

Dynamics of Circular Motion — Banking, Conical Pendulum and the Vertical Circle

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The Centripetal Force and What Supplies It

Centripetal force

F=mv2r=mω2rF = \frac{mv^2}{r} = m\omega^2 r

Forces at an Angle: Banked Roads, Conical Pendulums and Funnels

A tilted force turning a body

tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}

The Vertical Circle

Vertical circle

Tbottom−Ttop=6mg,vtop,min=gr,vbottom,min=5grT_{\text{bottom}} - T_{\text{top}} = 6mg, \qquad v_{\text{top,min}} = \sqrt{gr}, \quad v_{\text{bottom,min}} = \sqrt{5gr}

Common traps

Adding percentages

Scaling is multiplicative: 1.2×1.22÷1.2=1.441.2 \times 1.2^2 \div 1.2 = 1.44, a 44% rise. Adding the three 20%s is how 12% and 14% get into the options.

Asking for the new radius, answering the increase

'The increase in the radius of curvature' for 20% more speed on a 20 m curve is 8.8 m; the new radius, 28.8 m, sits beside it in the options.

Taking the bottom tension as 5mg

At the bottom T=mg+mv2rT = mg + \frac{mv^2}{r} and v2=5grv^2 = 5gr when just looping, so T=6mgT = 6mg. The 5mg5mg forgets the weight.

Moment of Inertia and Radius of Gyration

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Moment of Inertia of Standard Bodies

Definition

I=∑miri2,Iring=MR2,  Idisc=12MR2,  Isphere=25MR2I = \sum m_i r_i^2, \qquad I_{\text{ring}} = MR^2,\; I_{\text{disc}} = \tfrac{1}{2}MR^2,\; I_{\text{sphere}} = \tfrac{2}{5}MR^2

Radius of gyration

I=Mk2  ⇒  k=IMI = Mk^2 \;\Rightarrow\; k = \sqrt{\frac{I}{M}}

Rebuilt Bodies: Recasting, Bending and Scaling

Scaling for the same material

loop: I∝R3,disc (same thickness): I∝R4\text{loop: } I \propto R^3, \qquad \text{disc (same thickness): } I \propto R^4

Common traps

'Same material' does not mean 'same radius'

A solid sphere and a hollow one of the same MASS and material cannot have the same radius: the hollow one is bigger, so Ih>IsI_h > I_s even more clearly than the standard formulas suggest.

Comparing I and reporting it as k

The disc-to-ring ratio of I is 1 : 2, but of k it is 1:21 : \sqrt{2} — k is a square root. Both ratios are offered.

Keeping the radius when the material fixes the mass

For loops of the same wire, I∝R3I \propto R^3, not R2R^2 — the bigger loop also has more wire. A ratio of 27 in I is 3 in radius; answering 27\sqrt{27} forgets the mass.

Parallel and Perpendicular Axis Theorems

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The Parallel-Axis Theorem

Parallel-axis theorem

I=Icm+Md2I = I_{\text{cm}} + Md^2

The Perpendicular-Axis Theorem

Perpendicular-axis theorem (lamina)

Iz=Ix+IyI_z = I_x + I_y

Composite Bodies: Add the Parts, Subtract the Holes

Adding about one axis

I=∑parts(Icm,i+midi2)  −  IremovedI = \sum_{\text{parts}} \left(I_{\text{cm},i} + m_i d_i^2\right) \;-\; I_{\text{removed}}

Where the Centre of Mass Lies

Centre of mass

xcm=∑mixi∑mix_{\text{cm}} = \frac{\sum m_ix_i}{\sum m_i}

Common traps

Shifting from an axis that is not through the centre of mass

From an edge to a point midway you cannot add M(R2)2M(\frac{R}{2})^2. Go back to the centre-of-mass axis first, then out to the new one.

Using it on a three-dimensional body

Iz=Ix+IyI_z = I_x + I_y needs every bit of mass to lie in the x–y plane. For a sphere or a cylinder it gives nonsense; use the standard results instead.

Forgetting a sphere's own moment of inertia

A sphere on the axis still has 25MR2\frac{2}{5}MR^2 about it; a sphere off the axis has that PLUS Md2Md^2. Treating either as a point mass drops a term and misses every option.

Measuring from the wrong end

The centre of mass is nearer the HEAVIER mass. 2 kg and 4 kg on a 9 m bar: 6 m from the 2 kg, which is 3 m from the 4 kg — the options offer both ends.

Torque, Angular Momentum and Its Conservation

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Torque: From a Force, and From a Change in Spin

Torque

τ⃗=r⃗×F⃗,τ=Iα\vec\tau = \vec r \times \vec F, \qquad \tau = I\alpha

Angular Momentum and Its Links to Energy and Force

Angular momentum and energy

L=Iω,K=L22IL = I\omega, \qquad K = \frac{L^2}{2I}

Conservation of Angular Momentum

No external torque

I1ω1=I2ω2,K∝1I at fixed LI_1\omega_1 = I_2\omega_2, \qquad K \propto \frac{1}{I}\ \text{at fixed } L

Common traps

Writing F × r

Torque is r⃗×F⃗\vec r \times \vec F and angular momentum r⃗×p⃗\vec r \times \vec p. The options swap the order in one term to reverse its sign.

Equal energy means equal L

At equal kinetic energy, L=2IK∝IL = \sqrt{2IK} \propto \sqrt{I}: moments of inertia I and 2I give 1:21 : \sqrt{2}, not 1 : 2.

Conserving kinetic energy

Setting 12I1ω12=12I2ω22\frac{1}{2}I_1\omega_1^2 = \frac{1}{2}I_2\omega_2^2 gives ω\omega changing as I\sqrt{I}, which matches a planted option. The conserved quantity is IωI\omega.

Rotational Kinetic Energy and Rolling Motion

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Rotational Kinetic Energy, and Rods That Swing or Fall

Rotational kinetic energy

Krot=12Iω2K_{\text{rot}} = \tfrac{1}{2}I\omega^2

How a Rolling Body's Energy Splits

Rolling energy

K=12mv2(1+k2R2)K = \tfrac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)

Rolling Down (and Up) an Incline

Rolling on an incline

a=gsin⁡θ1+k2/R2,v=2gh1+k2/R2a = \frac{g\sin\theta}{1 + k^2/R^2}, \qquad v = \sqrt{\frac{2gh}{1 + k^2/R^2}}

Common traps

Raising the centre of mass by the whole length

A rod's weight acts at its middle, so falling from upright lowers it by L2\frac{L}{2}, not LL. Using LL gives 6g/L\sqrt{6g/L}, a printed distractor.

Forgetting the rotational part

A rolling body at speed v has MORE than 12mv2\frac{1}{2}mv^2. Setting mgh=12mv2mgh = \frac{1}{2}mv^2 for a rolling body is the sliding answer.

Using sin θ twice, or not at all

aa carries sin⁡θ\sin\theta; vv at the bottom depends only on the height hh. '30°, solid sphere' is 5g14\frac{5g}{14}, and 5g7\frac{5g}{7} is what you get by forgetting the sin⁡30∘\sin 30^\circ.

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