PHYSICS 36LOCUS

In terms of moment of inertia, equation (7.18) can be written as...(7.19 a)

The vector form of the above equation is

...(7.19 b)The angular acceleration produced is along the direction of applied net external torque. The magnitude of theproduced angular acceleration is directly proportional to the magnitude of the net external torque and inverselyproportional to the moment of inertia of the body. The above relation looks the translational equation net .F maHere, you should not forget that I is not an independent rule. It is derived from F = ma only. We can establishan analogue between translational and rotational variables. By doing so concept developed so far for translationalmotion would help to solve the problems involving rotational motion. The possible analogue is as follows

/ /( ) distance traversed, angle turned,

s( ) average speed, average angular speed,t t

( ) intantaneous speed, instantaneous a

i S

ii v

dsiii vdt

ngular speed,

( ) average acceleration, average angular acceleration,

( ) instantaneous acceleration, instantaneous angular acceleration,

( ) mass, moment of inertia, (

ddt

dviv adt t

dv dv adt dt

vi m Iv ) force, torque,

( )ii F

viii F ma I( ) linear momentum, Angular momentum, ( )

( )

( ) conservation of linear momentum: conservation of angular momentum:When 0, constant. When 0, constant.translational ki( )

ix p lx p mv l I

dp dlxi Fdt dt

xiiF p l

xiii2 21 1

2 2

netic energy, rotational kinetic energy,

( ) work done, work done,

k mv k I

xiv d F ds d d

Angular quantities involved in analogues (ix) to (xiv) would be discussed later in this topic.

Find the angular acceleration of the rod given in example 4 at the moment

(a) when it is released from rest in the horizontal position;

(b) when it makes an angle with the horizontal.

PHYSICS 37LOCUS

After the moment when the rod is released from the restin the horizontal position , it would rotate in the vertical plane abouta horizontal axis passing through the hinge and perpendicular to thelength of the rod. Initial angular velocity of the rod is zero but dueto nonzero torque of gravity it has some angular acceleration andhence, it will acquire some angular speed as it rotates. Asdiscussed in example 4, the hinge force does not provide anytorque about the axis under consideration and the weight of the rod tries to rotate it in the clockwise sense, i.e.,it provides a torque perpendicularly, inward to the plane of the paper. An approach using r F to find the torquewould also give the same result. Hence, angular acceleration of the rod,

gravitynet

I I

22

3mg lml

32

gl

(b) When the rod makes an angle with the horizontal, its angular acceleration,

mgnet

I I

mg rI

/

2

cos2

3

lmg

ml[from figure 7.40 (b)]

3 cos2

gl

In the previous case, find the angular velocity of the rod when it has turned through an angle after the moment whenit was released from rest in the horizontal position. Also find the angular velocity when the rod becomes vertical.

From the result obtained in part (b) of the previous example, at some angle , the angular acceleration ofthe rod,

3 cos2

gl

3 cos2

d gdt l

3 cos2

d d gd dt l [Using chain rule.]

0 0

3. cos2

gd dl at = 0, = 0

PHYSICS 38LOCUS

2

00

3sin

2 2gl

2 3 sin2 2

gl

3 singl

When the rod becomes vertical, 2, and hence, angular velocity,

3g l

In the previous example, find the hinge force on the rod at = 0.

Just after the moment when the rod was released from the rest in the horizontal position, it is shown infigure 7.41(a). Let the vertical component of the force on the rod from the hinge be 1R and the horizontal componentof the same be 2R , as shown in figure. The subsequent motion of the centre of mass of the rod is a nonuniformcircular notice on the vertical circular path of radius l/2 with the centre at the hinge, as suggested in the figure.

Initially the rod is at rest and hence radial component of the acceleration of the centre of mass of the rod, r, is zero.Hence, applying net cmF ma along the radial direction, we get

net, radial cm, radialF ma

2 0R m0

Applying the same along the tangential direction, we have, = .

net, tangential cm, tangentialF ma

1 2lmg R m

1 2lR mg m

34

mg mg

14

mg

Net hinge force 2 21 21 .4

R R mg 2[ 0]R

PHYSICS 39LOCUS

In the previous example, find the magnitude of the net hinge force on the rod when the rod has turned through anangle .

If be the angular velocity of the rod when it has turned through an angle , the centre of mass of the rod

has 2 2l

and 2l

as radial and tangential components of its acceleration, respectively, as shown in figure 7.41(b).

Applying ext cmF Ma on the rod along the radial direction, we have,

22 sin 2

lR mg m

23sin sin2

R mg mg

.

3 sin

3 cos2

gl

gl

=

=

Rod when it makes an angle with thehorizontal. , are perpendicular and radial components, respectively, of the reaction force acting

R R

on the rod from the hinge.

You can also assume reaction force as andacting at some angle with the rod.

R

5 sin2

mg

Applying the same along the tangential direction, we have,

1cos 2lmg R m

1 cos 2lR mg m

3cos cos4

mg mg

1 cos4

mg

Therefore, net force on the rod from the hinge can be obtained by solving2 21 2R R R

If the disc given in example 5 has mass M and it is free to rotate about its symmetrical axis passing through O, findits angular acceleration.

If be the angular acceleration of the disc, then, using net ,I we have,

net

I

23

( 2)FR

MR

6FMR

As the net torque is in clockwise sense, has the same sense of rotation.

PHYSICS 40LOCUS

A uniform disc of radius 0.12 m and mass 5 kg is pivoted so that itrotates freely about its axis. A thin, massless and inextensible stringwrapped around the disc is pulled with a force of 20 N, as shownin figure 7.42(a) .

(a) What is the torque exerted on the disc about its axis? (b) What is the angular acceleration of the disc?(c) If the disc starts from rest, what is the angular velocity after 3s?

It is obvious that the string force gives a torque to the disc in the clockwise direction. As the torque givenby the force from the axle is zero. Net torque on the disc is,

net torque of the string force

F rF R(20 N) (0.12 m)2.4 N-m.

As the net torque on the disc is in clockwise direction, the disc has angular acceleration in the same direction. If be the magnitude of the angular acceleration,

net net2( 2)I MR

22

2.4 2 rad/s5 (0.12)

266.66 rad/sAt t = 0 if the disc has zero angular velocity, then, at some time t, its angular velocity,

in0

t

t

At t = 3 s,

200 rad/s

A uniform disc of radius R and mass M is mounted on an axissupported in fixed frictionless bearing. A light string is wrappedaround the rim of the disc and a body of mass m is supported bythe string, as shown in figure 7.43(a).

(a) find the angular acceleration of the disc;(b) find the magnitude of the tangential acceleration of

the point on the rim where the string separatesfrom the rim.

(c) if the system is released from rest at t = 0, find thespeed of the block at some time t (>0).

PHYSICS 41LOCUS

Analyze the situation according to the information provided in the figure 7.43(b). You should also note thefollowing points: Only tension force of the string, T, produces a torque on the disc about its centre O. Torque of the weight of

the disc and that of the reaction force from the bearing are zero about O. If be the angular acceleration of the disc (in the clockwise direction) then the point P on the disc has a

tangential acceleration R in the vertically downward direction at the moment shown in figure. The stringunwinds at the same acceleration and the block has the same acceleration in the vertically downwarddirection. Therefore, if a be the acceleration of the block, then,

a R ...(i)Now, applying net I on the disc about its symmetrical axis, we have,

net I

2.

2MRT R

2M RT

2MaT [Using (i) ...(ii)

Using netF ma for the block in the vertical direction, we have,

netF ma

mg T ma ...(iii)Adding (ii) and (iii), we get,

2Mmg m a

2ma g R

m MIf v be the speed of the block at some time t, then, we have,

v u at is constanta

at 0u

2m gt

m M

Find the acceleration of 1m and 2m in an Atwoods Machine,shown in figure 7.44(a), if there is friction present between thesurface of pulley and the thread does not slip over the surface ofthe pulley. Moment of inertia of the pulley about its symmetricalaxis is I and its radius is R. The pulley can rotate freely about itssymmetrical axis.

PHYSICS 42LOCUS

Due to friction between the pulley and the thread tensions in the parts of the thread on the two sides of thepulley are different. Let that in the right part it is 1T and that in the left part is 2 ,T as shown in figure 7.44(b). Forcesacting on the two blocks and the pulley are also shown in figure 7.44(b). Force on the pulley from the support andits weight are not shown because they do not produce torque on the pulley about its symmetrical axis of rotation. Ifthe block 1m comes down with an acceleration a then 2m would go up with the same acceleration because theyare connected by the same string, as shown in the same figure.7.44(b).

If we assume that the pulley gets an angular acceleration in the clockwise sense then the torque of 1T would bepositive and that of 2T would be negative, as suggested in figure 7.44(c).Again, as any point on the rim of the pulley has a tangential acceleration R, the block 1m comes down and theblock 2m goes up with the same acceleration, as shown in figure 7.44(d).Therefore, we can write,

a R ...(i)

Using netF ma for the two blocks, we have,

1 1 1m g T m a ...(ii) [for 1m ]

2 2 2T m g m a ...(iii) [for 2m ]

Using net I for the pulley, we have,

1 2T R T R I ...(iv)torque of support forceand weight are zero

Substituting from (i) in (iv), we get,

1 2 2aT T I

R...(v)

Adding (ii), (iii) and (v), we get,

1 2 1 2 2( )Im m g m m a

R

1 22

1 2

m ma gm m I R

PHYSICS 43LOCUS

A thin uniform rod AB of mass m = 1.0 kg moves translationallywith acceleration a = 2.0 m/s due to two antiparallel force 1F and

2F acting on it perpendicularly to its length, as shown in figure7.45. The distance between the points at which these forces areapplied is x = 20 cm. Besides, it is known that 2 5.0 N.F Findthe length of the rod.

Before analyzing the details of the given situation, let us analyze the rotational effect of two antiparallelforces. Consider the situations shown in figure 7.45.

and are producing torques about in opposite directions.

F FA

21 and are producing torques about in opposite directions.

F FB

21 and are producing torques about in the same direction

F FC

21

If we analyze the torques of the two forces about every point in their plane containing them, then, we arrive at theconclusion that if the point lies between the lines of action of 1F and 2F then torques of the forces about that pointadd up together otherwise they are in opposite directions.If the magnitudes of the two forces are equal then such a pair is called as a couple. If the magnitude of each force isF and the distance between their lines of application is d, then, the net torque about any point in their plane is F.d,as shown in figure 7.47.

net 1 2

( )F l d FlF d

net 2 1

( )F l d FlF d

net 1 2

1 1 2 2

1 2( )F d F dF d dF d

Torque of a couple.

PHYSICS 44LOCUS

Now, let us discuss the given case. As the rod is in pure translationmotion, net torque on it about any point must be zero. Therefore,the centre of mass of rod can not lie between the lines of action ofthe forces because in that case torques produced by then aboutthe centre of mass do not cancel each other.

Let us assume that the centre of mass of the rod lies at a distance yaway from the line of action of 2 ,F as shown in figure 4.48. As therod translates towards right, 2F must have a greater magnitudethan 1.FUsing net ,F ma we have

2 1F F ma

1 2F F ma(5 1 2) N

= 3 NAgain, as the net torque on the rod about C must be zero,

1 2

magnitude of the torque magnitude of the torqueproduced by about produced by about F C F C

the two torques haveopposite directions

1 2( )F x y F y

2 1 1( )F F y F x

1

2 1

Fy xF F

3 205 3

cm

= 30 cmLength of the rod, 2( ) 1.0 ml x y

A force F Ai Bj is applied to a point whose radius vector relative to the origin of coordinates O is equal to ,r ai bj where a, b, A, B are constants, and ,i j are the unit vectors of the x and y axes. Find the torque

and the arm length l of the force F relative to the point O.

Torque of F about O is

r F ( ) ( )ai bj Ai Bj

( )aB bA kArm length of F with respect to O is

sinl r is the distance of the point of application of

from and is the angle between and .r F

O r F

r Fr sinr F r F

PHYSICS 45LOCUS

2 22 2 2

aB bAa b

a b A B

2

aB bA

A B

A uniform cylinder of radius R is spun about its axis to the angularvelocity 0 and then placed into a corner, as shown in figure 6.50(a).The coefficient of kinetic friction between the corner walls and thecylinder is equal to k. How many turns will the cylinder accomplishbefore it stops?

All forces acting on the cylinder are shown in figure6.50(b). As the cylinder rotates, its surface slips over the cornerwalls and hence frictional forces acting on it, 1f and 2 ,f are kineticin nature. Normal contact forces acting on the cylinder from thecorner walls, 1N and 2 ,N and the weight of the cylinder, mg,pass through the centre of the cylinder and hence, these forcesproduce no torque about the centre C. Only frictional forcesproduce torque about C and the torques produced by them are inopposite direction of the direction of the angular velocity of thecylinder and hence, they retard the rotational motion of the cylinder

As the cylinder does not translate, net force on it in both vertical and horizontal directions must be zero. Therefore,

1 2N f mg

1 2N N mg ...(i) 2 2[ ]f Nand 2 1N f

2 1N N ...(ii) 1 1f NSubstituting for 2N in equation (i) from equation (ii), we have,

21 1N N mg

1 21mgN ...(iii)

Substituting for 1N in equation (ii) from equation (iii), we have,

2 21mgN ...(iv)

If we define the anticlock wise sense of rotation as the +ve direction of rotation, then, the clock wise sense becomesthe ve direction for the same. Hence, angular acceleration in the present case becomes negative for this choice ofrefrence direction. The angular acceleration,

1 2net torque due to and about moment of inertia about the axis of rotation

f f C

PHYSICS 46LOCUS

1 22

2

f R f RmR

1 22( )N NmR

1 22 ( )N NmR

22 1

1mg

mR

22 1

1g

R

If the cylinder had the angular velocity 0 at t = 0and at some time t it has an angular velocity , and in this durationit has turned through an angle , then,

2 20 2

20 2

4 11

gR

If the cylinder stops having rotated through an angle 0 , then at 0 , = 0. Therefore,

2 20 02

4 101

gR

2 20

0(1 )

4 (1 )Rg

Therefore, the number of rotations accomplished by the cylinder, before it stops,

0

2n

2 20 (1 )

8 (1 )Rg