In the section ** ‘Circular motion and rotation’**, the sub sections included as common for AP Physics B and C are

**(1) Uniform circular motion **and

**(2) Torque and rotational** **statics**.

The sub sections included for AP Physics C only are

**(1) Rotational kinematics and dynamics** and

**(2)** **Angular momentum and its conservation**. ** **

*Circular motion and rotation’* carries 4% of the total points for AP Physics B and 9% of the total points for AP Physics C.

Here are the important equations you need to remember to tackle the AP Physics B Examination:

**(1**) When an object is in ** uniform circular motion**, the magnitude of its

*centripetal acceleration***(**is given by

*a*)_{c}* a _{c }*

**=**where ‘

*v*^{2}/R*v*’ is the speed of the object and

*R*is the radius of the circle. The direction of

*a*is always towards the centre of the circle.

_{c}The ** angular speed ω** is related to the speed

*v*by

** v = ωR**.

Therefore, the ** centripetal acceleration** is

*a*=_{c}*ω*^{2}*R*.If *T *is the time period of revolution of the object in circular motion and *f *is its frequency, we have *f *= 1/*T** * and *ω*= 2π/*T** *= **2π f.**

The **centripetal force **required to produce *circular motion* is ** mv^{2}/R = ** m

*ω*^{2}*R***(2) Angular acceleration ( α)** is the time rate of change of angular velocity and is given by

*α* = d*ω*/dt

In the case of a body in accelerated rotational motion, the angular velocity (*ω*) after a time *t* is given by

*ω = ω*_{0}** + ***αt*** **where *ω*_{0 }is the initial angular velocity and and *α* is the angular acceleration. [This is similar to the equation, *v = v*_{0} + *at* in linear motion].

The ** angular displacement (θ) ** at the instant

*t*is given by

*θ = θ*_{0} + *ω*_{0}*t*** + (½) ***αt*** ^{2}** where

*θ*

_{0}is the initial displacement (at

*t*=0)

[This is similar to the equation x = *x*_{0} + *v*_{0}*t* + (½) *at*^{2} in linear motion].

**(3) Torque (**

**τ)**is the moment of force and is the

**. By lever arm we mean the perpendicular distance (ON in the figure) of the line of action of the force from the origin.**

*product of force and the lever arm*Therefore, Torque τ produced by the force F acting on the particle at P = ON × F.

Since ON = r sinθ where θ is the angle between** r** and **F**,

τ = rF sinθ

[Note that torque (**τ**)** **is a vector which is the vector product of the position vector** r **and the force vector** F**. Therefore, **torque ****τ** = **r **× **F**]

**(4) **A** **rigid body will be in mechanical equilibrium, if the total force and the total torque on the body are zero. The condition of zero net force will ensure that there is *no change* in the *linear momentum* and the condition of zero net torque will ensure that there is *no change* in the *angular momentum*.

For **AP Physics C** Examination, you will require the following also (in addition to the above):

**(5) **Consider a particle of mass *m *and linear momentum **p**. If the position vector of the particle is **r**,** **the **angular momentum** **L **of the particle with respect to the origin is given by

** L **= **r **× **p**

**L **is a vector whose magnitude is *rp*sinθ where *r* is the distance of the particle from the origin, *p* is the magnitude of the linear momentum vector **p** and θ is the angle between the vectors **r **and **p**.

Angular momentum is the moment of the linear momentum and is the product of linear momentum and the lever arm *r *sinθ. [By lever arm we mean the perpendicular distance of the line of action of the linear momentum from the origin].

**(6) Moment of inertia ( I)**of a system of particles (as in the case of a rigid body) about an axis (of rotation) is given by

*I* = ∑*mr*^{2 }where *m* is the mass of a particle at perpendicular distance *r* from the axis of rotation and the summation is for all the particles.

*Radius of gyration ***( k) **is related to moment of inertia (

*I*) and the mass of the body (

*M*) as

*I = Mk*^{2}

Note that both *I *and* k* depend on the axis of rotation.

** Parallel axes theorem **states that the moment of inertia (

*I*)of a body about any axis is equal to the sum of the moment of inertia (

*I*

_{CM})of the body about a parallel axis through its centre of mass and the product

*Ma*

^{2}where

*M*is the mass of the body and

*a*is the distance between the two axes:

*I = I*_{CM}* + Ma*^{2}

** Perprndicular axes theorem** states that the moment of inertia of a lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.

[For example, if the X and Y axes are in the plane of the lamina, the Z-axis is the perpendicular axis and we have ** I_{Z}** =

**where**

*I*_{X}+*I*_{Y}

*I*

_{Z},

*I*

_{X}and

*I*

_{Y}are the moments of inertia about the Z, X and Y-axes respectively]

** Moments of inertia of some regular bodies** are given below:

**(i)** **Thin circular ring** (Mass *M*, Radius *R*) about its central axis perpendicular to its plane: *MR*^{2}

** A hollow cylinder (pipe) **also has the above value for its moment of inertia about its own axis.

**(ii) Thin circular ring** (Mass *M*, Radius *R*) about any diameter: *MR*^{2}/2

**(iii) Thin rod **(Mass *M*, Length L) about a perpendicular axis through the mid point: *ML*^{2}/12

**(iv) Circular disc **(Mass *M*, Radius *R*) about its central axis perpendicular to its plane: *MR*^{2}/2

**(v) Circular disc **(Mass *M*, Radius *R*) about its diameter: *MR*^{2}/4

**(vi) Solid cy**l**inder **(Mass *M*, Radius *R*) about the axis of the cylinder: *MR*^{2}/2

**(vii) Solid sphere **(Mass *M*, Radius *R*) about its diameter: (**2/5) MR^{2}**

**(viii) Hollow sphere **(Mass *M*, Radius *R*) about its diameter: (**2/3) MR^{2}**

**(7) Angular momentum ( L)** is given by

** L = Iω** where

*I*is the moment of inertial about the axis of rotation and

*ω*is the angular velocity.

This is similar to the expression for linear momentum ** p = mv**. In angular motion (rotational motion)

*I*is to be used in place of

*m*and

*ω*is to be used in place of v.

**Newton’s 2 ^{nd} law** in rotational motion is

**τ = d L/dt = d(**

*Iω***)**

*/dt*If the moment of inertia (*I*)* *is constant, as is the case of a rigid body rotating about a fixed axis, we can write

**τ = ***I ***( dω/dt) = I**

**where**

*α**α*is the angular acceleration.

The **law of conservation of angular momentum** states that in the absence of external torque, the angular momentum of a system remains unchanged. This can be expressed as

** I_{1 }ω_{1} = I_{2} ω_{2}** where

*I*

_{1}and

*I*

_{2}are the initial and final moments of inertia and

*ω*

_{1}and

*ω*

_{2}are the initial and final angular velocities of a system in the absence of external torques.

**(8) Acceleration ( a) of a body rolling down an inclined plane **of inclination θ is given by

** a = gsinθ / [1 + (k^{2}/R^{2})]** where

*R*is the radius of the body and

*k*is the radius of gyration

**about the axis of rolling.**

Since *Mk*^{2} = (2/5)*MR*^{2} for a solid sphere, *k*^{2}*/R ^{2}* = 2/5. This is the

*least value*in the case of regular bodies and hence the acceleration

*a*is maximum in the case of a solid sphere. In contrast, in the case of a ring (and pipe), the value of

*k*

^{2}

*/R*is 1 and is the maximum in the case of regular bodies and hence the acceleration

^{2}*a*is the minimum in the case of a ring.

If differently shaped bodies are allowed to roll down from the top of an inclined plane, the *solid sphere* will reach the bottom *first *and the *ring* (and the pipe) will arrive *last*. It is interesting to note that for a given shape, the time of arrival at the bottom is *independent of mas and size.*

** **The above equation** **for acceleration down the plane can also be written as

** a = (Mgsinθ) / [M + (I/R^{2})]** where

*M*is the mass of the body; but it will be better to remember the above form in terms of the radius of gyration,

*k*.

**(9) Rotational kinetic energy ( K) of a body** is given by

*K* = (½) *Iω*^{2}

**(10) A rolling body has translational and rotational kinetic energies. **The total kinetic energy of a rolling body is therefore given by

** K = (½) Mv^{2 }+ (½) Iω^{2} **where

*v*is the linear velocity of the body

**(11) Work (d W) done by a torque **

**τ**in producing an angular displacement d

*θ*is given by

**d W =**

**τ**[By d

*d*θ*W*we mean the

*small*amount of work done for a

*small*angular displacement d

*θ*]

This is similar to the expression for work, d*W *= Fds in linear motion.

**(12) Power ( P) in rotational motion **is given by

** ***P ***= τω**

This is similar to the expression for power, *P *= *Fv* in linear motion

**Different ****quantities in linear motion**** and the corresponding ****quantities in rotational motion**** (about a fixed axis) are given below:**

(a) Displacement *x *→ Angular displacement θ

(b) Velocity *v *= d*x*/d*t* or d*r*/d*t*→* *Angular velocity ω = dθ/d*t*

(c) Acceleration *a *= d*v*/d*t *→* *Angular acceleration α = dω*/*d*t*

(d) Mass *M*→* *Moment of inertia *I*

(e) Linear momentum *p *= *Mv*→* *Angular momentum *L *= *I*ω

(f) Force *F *= *Ma *→Torque τ = *I *α

(g) Work *dW *= *F*d*s *→* *Work d*W *= τ *d*θ

(h) Kinetic energy *K *= *Mv*^{2}/2→ Kinetic energy *K *= *I*ω^{2}/2

(i) Power *P *= *F v*→* *Power *P *= τω

In the next post, we will discus typical questions in this section.

Allright,I understood the formula of calculating the acceleration of a rolling body down the formula,is there any formula to calculate the velocity of a rolling body down the inclined plane.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteHello roshanboy,

ReplyDeleteTo obtain the linear velocity down the plane, just substitute the value of the acceleration in the equation of linear motion such as v = u + at or v^2 = u^2 + 2as. You can use the energy relation, mgh = (1/2) mv^2 +(1/2)Iω2 also to find the velocity. Here ‘h’ is the height of the inclined plane, ‘I’ is the moment of inertia and ‘ω’ is the angular velocity which is v/R where R is the radius of the rolling body. Don’t think of a ready made formula for ‘v’.