Rotation of a Rigid Body

1 Rotation of a Rigid Body

Rotation of a Rigid Body1. A wheel is rotating at the rate of 50 rev/min in the anti clockwise direction What should be the magnitude and direction of the angular acceleration of the wheel so that it stops in 8 s? How many revolutions will it cover before stopping?Solution:Let ωi=¿ initial angular velocityωi=¿

here ,ωf=0

α=ωf−ωi

t=0−5 π /3

8=−5π24

rad /s2

Hence angular acceleration of 5 π /24 rad / s2 must be imparted to the clock in the clock-wise direction.Angular displacement of θ is given as:θ=

ωf2−ωi

2

2α=0−(5 π /3)2

−2¿¿

⇒θ=20π3

radians

Number of accelerations 02π

=103

revolutions anti clockwiseHence it completes (3+1/3) revolutions before stopping.2. A flywheel of radius 30 cm starts from rest and accelerates with constant acceleration of 0.5 rad / s2 . compute the tangential, radial and resultant accelerations of a point on its circumference:

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(a) Initially at θ=0 °∧¿ (b) After it has made one third of a revolution.

Solution:(a) At the start:α=0.5 rad /s2

R=0.3m

ω=ωi=0 rad / s

radial accleration=ar=ω2R=0m /s

tangential acceleration=ar=Rα

¿ (0.3 ) (0.5 )=0.15m /s2

Net acceleration=anet

¿√at2+a t

2=√02+0.152=0.15m/ s2

(b ) Afterθ=120 °(2π /3)

ωf2=ωi

2=2αθ=0+2 (0.5 )(2 π /3)

⇒ωf=√ 2 π3 rad / s

a t=ω2R=2π /3 (0.3 )=π /5m /s2

a t=Rα=(0.3 ) (0.5 )=0.15m /s2

anet=√at2+at

2=√ π2

25+ (0.15 )2

¿0.646m / s23. A wheel mounted on a stationery axle starts at rest and is given the following angular acceleration:α=9−12 t (¿ SI unit s)where t is the time begins to rotate. Find the number of revolutions that the wheel turns before it stops (and begins to turn in the opposite direction).Solution:The kinematic equations do not apply because the angular acceleration α is not constant.We start with the basic definition: α=dx /dt ¿write ¿



ω−ω0=∫0

t

α dt=¿∫0

t

(9−12 t )dt=9 t−6 t2 ¿

(in SI units)We find the elapsed time t betweenω0=0∧ω=0by subsituting these values :

0−0=9 t−6 t2

Solving for t, we obtain t=9 /6=1.50 sθ−θ0

¿∫0

t

ωdt=∫0

t

(9 t−6 t f2) dt=4.5 t 2−2t 2

Subsitutingθ0=0∧1.5 s ,weobtain

θ−0=4.5 (1.5 )2−2 (1.5 )3=3.375 rad

4. Find the moment of inertia of a circular disk or solid cylinder of radius R about the axis through the corner and perpendicular to the flat surface.Solution:The figure shows that the appropriate mass element is a circular ring of radius r and width dr.Its area is dA=2 π dr∧its massis dm=σ dA ;

where σ=M /A isthe areal massdensity .

The moment of inertia of this element is:dI=dmr2=2π σ r3ar

For the whole body,



I=2πσ∫0

R

r3dr=12πσ R4

The mass of the whole disk or cylinder is M¿σA=σπ R2 ,∧so

I=1/2M R2

5. Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter.Solution:The sphere may be divided into disks perpendicular to the given axis, as shown in fig. the disk at a distance x from the center of the sphere has a radius r=(R2+x2)1/2 and a thickness dx.If ρ=M /N is the volume mass density (mass per unit volume), the mass of this elemental disk is dm= ρdV =ρπ r2dx ,∨¿

dm= ρπ (R2+x2 )dx

From the last example, the moment of inertia of this elemental disk isdt=1/2dm r2=1/2 ρπ (R2−x2)2dx

The total moment of inertia isI=1/2 ρπ∫

R

R

(R4−2 R2 x2+x4 )dx

¿1/2 ρπ [R4 x−23 R2 x3+ 15x5]

R

R

= 815

ρπ R3

The total mass of the sphere is M = ρ ¿ so the moment of inertia may be written as:Edudigm 1B Panditya Road Kolkata-29|033 4003 4819|Edudigm.in


I=2/5M R2 .

6. A grind stone is in the form of a solid cylinder has a radius of 0.5 m and a mass 50 kg.(a) What torque will bring it from rest to an angular velocity of 300 rev/min in 10 s?

(b) What is the kinetic energy when it is rotating at 300 rev/min?Solution:Letωi=0 rad / s ,ωf=2π ¿¿

(a )α=ωf−ωi

t=10π−0

10=π rad /s2

Torque required ¿ τ=I α=(12 M R2)α

⇒ τ=1/2 (50 ) (0.5 )2π=19.6N m

(b) Kinetic energy of a rotating body (RKE):RKE=1

2I ω2=1

2 ( 12 M R2)ω2

¿ 12 ( 12 (50 ) (0.5 )2) (10π )2

R K E = 3084 J7. Calculate the torque developed by an airplane engine whose output is 2000 hp at an angular velocity of 2400 rev/min.Solution:ω=2π ¿

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¿ ( tor que )×(angular dispacement )

Power=work done per sec=τ=∆θ∆ t

power=τ ω

τ=2000×74680π

=5937N m

8. In the given figure, calculate the linear acceleration of the blocks.Mass of block A = 10 kgMass of block B= 8 kgMass of disc shaped pulley= 2 kg¿

Solution:Let R be the radius of the pulley and T 1∧T2 be the tensions in the left and right portions of the string.Letm1=10kg ;m2=8kg ;M=2kg

Let a be the acceleration of blocks.

For the blocks (linear motion)(i )T 1−m2g=m1a

(ii )m2g−T2=m2a



For the pulley (rotation)net torque=Iα

( iii )T 2R−T 1R=12M R2α

The linear acceleration of blocks is same as the tangential acceleration of any point on the circumference of the pulley which is Rα .

( iv )a=Rα

Dividing (iii) by R and adding to (i) and (ii),m2g−m1g=m2a+m1a+

M2

Rα

⇒m2g−m1g=(m2+m1+M2 )a

a=m2−m1

m2+m1+M2

¿(10−8 )g

10+8+22

=2019

m/ s2 .

9. A uniform rod of length L and mass m is pivoted freely at one end.(a) What is the angular acceleration of the rod when it is at angle θ to the vertical?(b) What is the tangential linear acceleration of the free end when the rod is horizontal? The moment of inertia of a rod about one end is 1/3M L2 .



Solution:

The figure shows the rod at an angle θ to the vertical.If we take torques about the pivot we need not be concerned with the force due to the pivot.The torque due to the weight is mgL /2 sinθ, so the second law for the rotational motion ismgL2

sinθ= M L2

3α

Thus ,α=3gsinθ2L

When the rod is horizontal θ=π /2∧α=3 g/2 L.

The tangential liner acceleration of the free end is a t=αL=3 g2

10.A rigid body of radius of gyration K and radius R rolls (without slipping) down a plane inclined at an angle θ with horizontal. Calculate its acceleration and the frictional force acting on it.Solution:When the body is placed on the inclined plane, it tries to slip down and hence a static friction f acts upwards. This friction provides a torque which causes the body to rotate.



Let a be the linear acceleration of center of mass and α b the angular acceleration of the body.

For rotation around the axis through center of massnet torque=Iα⇒ fR=(mk2 )α

As there is no slipping, the point of contact of the body with plane is instantaneously at rest.⇒ v=Rω∧a=Rα

Solve the following three equations for a and f:mg sinθ−f =ma

fR=mk2α

a=Rα

a=g sinθ

1+ k2

R2 ¿

f=mg sinθ

1+ R2

k 2

We cal also derive the condition for pure rolling. (rolling without slipping)¿avoid slipping f ≤μgN

gsinθ

1+R2/k2≤ μgmg cosθ



⇒ μg≥tanθ

1+ R2

k2

This is the condition on μg so that the body rolls without slipping.11. A solid cylinder rolls down an inclined plane of height h and inclination θ . calculate its speed at the bottom a, the plane using acceleration-method and energy-method. Also calculate the time taken to reach bottom.Solution:Energy Method:Let v, ω be the velocity of center of mass and the angular velocity of cylinder respectively at the bottom of the plane.

As the cylinder rolls down:Loss in G.P.E= gain in translation KE+ gain in rotational KEmgh=1

2mv2+ 1

2I ω2

As the cylinder is rolling without slippingv=Rω

⇒mgh=12mv2+ 1

2¿

⇒ v=√ 4 gh3Acceleration Method:From the result of last example ¿Acceleration of the cylinder is;



g sinθ

1+12

=23g sin θ

Using v2=u2+2 as down the plane (taking downward direction positive)⇒ v=√0+2a h

sinθ

¿√2( 23 gsinθ) hsinθ

= √ 4 gh3Time to reach bottom = t¿ v−u

a=√√ 4 gh3 −0

23gsinθ

¿√ 3hg .1

sinθ

12. A solid cylinder of mass m = 4 kg and radius R= 10 cm has two ropes wrapped around it, one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the ceiling such that both the cords are exactly vertical. The cylinder is released to fall under gravity. Find the tension in the cords when they unwina and the linear acceleration of the cylinder. Solution:Let a = linear acceleration andα=¿ angular acceleration of the cylinder.For the linear motion of the cylinder:mg−2T=ma

For the rotational motion:Edudigm 1B Panditya Road Kolkata-29|033 4003 4819|Edudigm.in


net torque=1α

2TR=( 12mR2)α

Also, the linear acceleration of cylinder is same as the tangential acceleration of any point on its surface.a=Rα

Combining the three equations we get:mg+ma+m

2a⇒ a=2g

3=6.53m/ s2∧¿¿

T=mg−ma2

=6.53N

13. The end A of the rod AB is being pulled on the floor with a constant velocity v0 as shown. Taking the length of the rod asl, Calculate:

(a) The velocity of end B(b) The angular velocity of the rod(c) The velocity of the CM of the rod at the instant when θ=37 ° .Solution:LetCA=x ,OB= y

x2+ y2=l2∧x=l cosθ

(a) Differentiate x2+ y2=l2 with respect to time2 x

dxdt

+2 yd ydt

=0

V B=dydt

=−xy

dxdt

=−V A cotθ=43V 0↓



(b )Differentiate x=l cosθ with respect to time dx

dt=−l sinθ=dθ

dt

dθdt

=−V A

l sinθ

(negative sign indicatesthat θ is decresing)

ω=|dθdt |= V 0

l sin 37 °=5v 03 l

(c ) xcm=x2

⇒V CM=12dxdt

=v02

(→)

ycm=y2

⇒V CMy=12dydt

=vB

2=23v0↓

⇒V CM=√ v02

4+4 v0

2

9

¿ 56v0 tan

−1 43below horizontal.

14. Consider the arrangements shown in figure. The string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a massless, frictionless pulley to a falling weight. Determine the acceleration of the falling mass m in terms of only the mass of the cylinder M, the mass m and g.Solution:



Let T be the tension in the string and f be the force of (static) friction between the cylinder and the surface.A= downward acceleration of block mA=acceleration of the CM of cylinder towards rightα=¿ angular acceleration of cylinder (clockwise)T+ f =MA (i)

Taking torque about the CM,Tr−fr=¿

mg−T=ma(iii)

The string attaches the mass m to the highest point of the cylinder.V m=V CM+rω(iv)

We also have: (for rolling without slipping)A=rα (v )

Multiply (i) by r, (iii) by 2r and add to (ii)2mgr=MA r+2mar+1/2Mr 2α

¿ iv∧v ,wehave a=2 r α , A=r α

2mgr=¿

α= 4mg(3M+8m) r

The acceleration of falling mass m is:a=2 r α= 8mg

3M+8m



15. A rectangular rigid fixed black has a long horizontal edge. A solid homogenous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in figure. There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine:(a) The angle of θ0through which the cylinder rotates before it leaves contact with the edge.(b) The speed of the center of mass of the cylinder before leaving contact with the edge, and(c) The ratio of the transitional and rotational kinetic energy of the cylinder when its center of mass is in horizontal line with the edge. Solution:As the cylinder rotates about the edge, its CM moves along a circular arc of radius r centered at the edge. Let N be the normal reaction from edge at the instant when cylinder has rotated through the an angle θ .mg cosθ−N=mv2

r

At jump off point, N=0⇒mg cosθ=mv2

r

We can also apply conservative of mechanical energy as there is no slipping. The gravitational potential energy decreases and hence the kinetic energy increases.Edudigm 1B Panditya Road Kolkata-29|033 4003 4819|Edudigm.in


mgr ¿

Combine all the three equations to get:θ=cos−1 4

7∧V cm=√ 47 rg

(c)The rotational kinetic energy ( 12 1cmω2) becomes constant after the cylinder leaves the edge.Knot=

121cmω

2=12 (12 Mr2) 47 g

r=Mgr

7

The kinetic energy at the instant when CM is in horizontal line with edge is K.K= loss in GPE from initial position= mgrK trans=mgr−Knot=mgr−mgr

7=67mgr

⇒K trans

K not

=6

16. A solid sphere of radius r and mass m rolls without slipping down the track shown in the figure. At the end of its run at point Q its center of mass velocity is directed upward.Solution:From A to BLoss in GPE = gain in KE

mg (10R )=12mV cm1

2 + 121cmω1

2

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V cm1=Rω1

The CM follows a circular path of radius R-rAt B, net force towards center ¿N−mg=

mV cm2

R−r

⇒N=mg+m(100gR)7(R−r ) =mg(107 R−7 r)

7 (R−r)

(b )¿ A ¿Q ,mg (9R+r )=12mvc m2

2 +121cm(V c m2

r )2

(V c m2=rw2

at Q)

From Q to P, w does not change because about CM torque in zero in air.Gain in GPE= loss in KE⇒mg×gain∈height=1

2mV cm2

2

⇒h=V c m2

2

2 g=57(9 R+r )

⇒height above the base=R+h=52 R7

+ 5 r7

17. The descending pulley (disc shaped) shown in figure have a radius 20 cm and moment of inertia 0.20kg−n t 2 . the fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.Solution:Let

T 1∧T2 be tension in the strings to the left

and right of the pulley.a= acceleration of block towards right



A= acceleration of CM of pulley downwardα=¿ angular acceleration of pulley (anticlock wise)T 1=ma…………(i)

Mg−T 1−T 2=MA………(ii)

Taking torque about CM of the pulleyT 2R−T1 R−1α ……(ii)

The string attaches the mass m to the left point of the cylinder.⇒ a=A+Rα……. (iv)

for disc , I=12M R2…….(v)

We also have: (for right part of the string)A=Rα…….(vi)

Solving these equations for a, we get: ¿ 4 lg

4m l2+9 I

⇒ a= 4×0.2×9.8

4×1× (0.2 )2+9×0.2=4m /s2

18. In the arrangement shown in figure a weight A possess mass m, a pulley B possesses mass M. also known are the moment of inertia I of the pulley to its axis and the radii of the pulley R and 2R. The mass of the threads is negligible. Find the acceleration of the weight a after the system is set free.Solution:T 1=¿ tension in string attached to ceilingT 2=¿ tension in string attached to block



Acm : acceleration of CM of spool downwardMg+T2−2T 1=M Acm……(i)

mg−T 2=ma…….(ii)

Taking torque about CM,2T1 (R )+T 2 (2 R )=Iα …….(iii)

For the string winding over, bigger part of the spoolAcm=Rα

And for the string winding over the smaller parts of spoola=Acm+2Rα

Solving these five equations for a, we get:a=

3g (M+3m)M+9m+ I /R2

19. A gridstone 1m is diameter, of mass 50 kg, is rotating at 900 rev/min. A tool is pressed normally against the rim with a force of 200 N, and the gridstone comes to rest in 10 seconds. Find the coefficient of friction between the tool and the gridstone. Neglect the friction in the bearings.Solution:Let μ=¿ coefficient friction between the tool and the grid stone.Initial angular velocity¿ω1=900 rev /min

¿ 900×2π60

¿30 πrad /sec(anticlockwise)



Final angular velocity ¿ωr=0

Taking torque about CMfR=Iα⇒ fR=1

2M R2α∧α= 2 f

MR=2 Fμ

MR(clockwise)

Usingωf=ωi+αt , wehave

⇒0−30 π+(−2 FμMR ) t

⇒ μ=30MRπ2Ft

¿ 30×50×0.5×3.142×200×10

=0.5890

20. A carpet of mass m made of an inextensible material is rolled along the its length in the form of a cylinder of radius R and is kept on a length on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2.Solution:As radius is reduced to R/2, mass of rolling part reduces to M/4Loss in GPE = gain in KE⇒ M4

gR2

+ 3M4

gR=12 ( M4 )V cm

2 + 121ω2……(i)

I= moment of inertia of rolling part¿ 12 ( M4 )( R2 )

2

=M R2

32……… ..(ii)

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V cm=R2ω….( iii)

Solving this equation, we get: V cm=√ 14 gl3


Rotation of a Rigid Body

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