Revision Note 8: Circular Motion

Correspondence between Circular and Linear quantities:

Quantity	Linear	Circular
Displacement	x	θ
Velocity	v = dx/dt	ω = dθ/dt
Acceleration	a = dv/dt	α = dω/dt
Mass/moment	m	I = Σmr2
Force/torque	F	Τ
2nd law	F = ma	Τ = Iα
Impulse	F.dt	Τ.dt
Momentum	p = mv	L = Iω
Work	dW = F.dx	dW = Τ.dθ
Power (P=dW/dt)	P = F.v	P = Τ.ω
Kinetic Energy	KEl = ½ mv2	KEθ = ½ Iω2

Moment of Inertia:

Moment of inertia is the circular equivalent of mass. It deserves special mention as it is smartly crafted to make the 2^nd law and energy relations turn out to be equivalent.

Unlike mass, moment of inertia doesn’t depend just on the body, but also on the axis of rotation. If r is the radius of rotation, then:

r ↑        I↑↑

Conservation of momentum:

Both linear and circular momentums are conserved.

But there is no concept of the sum of linear and angular momentums being conserved – the two have different dimensions and cannot be added.

Kinetic Energy:

Both KE_l  and KE_θ have the same dimension. Suppose a disk (of mass m) rolls downhill through a vertical drop of h. Then the lost potential energy (mgh) is transferred partially to KE_l  and partially to KE_θ. So,

mgh = KE_l + KE_θ

Further solution – knowing the values of KE_l  and KE_θ – needs more equations.

Radial and Tangential acceleration:

Tangential acceleration = αr

Radial acceleration = ω²r

In the case when there is no angular acceleration, tangential acceleration is 0.

Parallel axis theorem:

I_p = I_cm + Md²

Problem Solving Tips:

Tip 8.1: Bridging Linear and Circular quantities

Most problems ask to relate the linear and circular quantities. The bridges are:

Kinematics Bridge:

x = θr

v = ωr

a = αr

Force-Toque Bridge:

F.r = T

Tip 8.2: Finding moment of inertia

General method is to take small pieces of mass, dm and integrate ½ r².dm. From the problem dm = K.dr, where K is some expression. No bridge is needed.

Tip 8.3: Disk rolling downhill

mgh = KE_l + KE_θ = ½ mv² + ½ Iω²

Then use the bridge: v = ωr

Tip 8.4: Massive pulley

If there are two strings, their tensions are different, and then the bridge equation is:

Torque = (T₂ – T₁).r

Sometimes the bridge v = ωr is needed.

Tip 8.5: Bullet embeds in a door

Initial angular momentum = angular momentum of the bullet = L = Iω = mr² . v/r = mvr.