1. CIRCULAR MOTION

Circular Motion

1. Calculate the angular velocity and linearvelocity of a tip of minute hand oflength 10 cm.

Given :T = 60 min. = 60 ×××× 60 s

= 3600 sl = 10 cm = 0.1 m

To Find :ωωωω = ?v = ?

Formula :

ωωωω =ππππ

T

2

v = rωωωωSolution :

ωωωω =ππππ

T

2

=××××2 3.142

3600

∴∴∴∴ ωωωω = 1.744 ×××× 10–3 rad/s

v = rωωωω

= 0.1 ×××× 1.745 ×××× 10–3

∴∴∴∴ v = 1.745 ×××× 10–4 m/s

2. Propeller blades in aeroplane are 2 mlongi) When propeller is rotating at

1800 rev/min, compute thetangential velocity of tip of theblade.

ii) What is the tangential velocity at apoint on blade midway between tipand axis ?

Given :r1

= 2 m,n = 1800 r.p.m

=1800

60 = 30 r.p.s

To Find :i) Tangential velocity, v

T1= ?

ii) Tangential velocity, vT1= ?

Formula :v = 2ππππnr

Solution :i) Tangential velocity of,

the tip of blade,vT1

= 2ππππnr.r1

= 2 ×××× 3.14 ×××× 30 ×××× 2∴∴∴∴ v

T1= 376.8 m/s

ii) Tangential velocity at a pointmidway between between tip andaxis,vT2

= 2ππππnr2

= 2 ×××× 3.14 ×××× 30 × × × × 1∴∴∴∴ v

T2= 188.4 m/s

3. A car of mass 2000 kg moves round acurve of radius 250 m at 90 km/hr.Compute :i) angular speedii) centripetal accelerationiii) centripetal force.

Given :m = 2000 kg,r = 250 m,v = 90 km/h

= 90 ××××5

18= 25 m/s

To Find :i) ωωωω = ?ii) a

cp= ?

iii) Fcp

= ?Formula :

i) ωωωω =r

v

ii) acp

= ωωωω2r

iii) Fcp

=mv

r

2

Solution :

ωωωω =25

250

∴∴∴∴ ωωωω = 0.1 rad/s

.. 2 MAHESH TUTORIALS SCIENCE

Circular Motion

4. A bucket containing water is whirledin a vertical circle at arms length. Findthe minimum speed at top to ensure thatno water spills out. Also findcorresponding angular speed.[Assume r = 0.75 m]

Given : r = 0.75 m, g = 9.8 m/s2

To Find :V

Top= ?

ωωωω = ?Formula :

VTop

= rg

ω ω ω ω =V

r

Top

Solution :

VTop

= ××××0.75 9.8

∴∴∴∴ VTop

= 2.711 m/s

ω ω ω ω =2.711

0.75

∴∴∴∴ ω ω ω ω = 3.615 rad/s

5. A motor cyclist at a speed of 5 m/s isdescribing a circle of radius 25 m. Findhis inclination with vertical. What is thevalue of coefficient of friction betweentyre and ground ?

Given :v = 5 m/s2

r = 25 mg = 9.8 m/s2

6. A stone weighing 1 kg is whirled in avertical circle at the end of a rope oflength 0.5 m.Find the tension ati) lowest positionii) mid positioniii) highest position

Given :m = 1 kgr = 0.5 mg = 9.8 m/s2

To Find :i) T

L= ?

ii) TM

= ?

iii) TH

= ?Formula :

i) TL

=mv

rL

2

+ mg

ii) TM

=mv

r

2m

iii) TH

=mv

–mgr

2H

acp

= ωωωω2r

acp

= (0.1)2 ×××× 250∴∴∴∴ a

cp= 2.5 m/s2

Fcp

=mv

r

2

Fcp

=( )××××2000 25

250

2

∴∴∴∴ Fcp

= 5000 N

To Find :θθθθ = ?µµµµ = ?

Formula :

tan θθθθ =v

rg

2

µ µ µ µ =v

rg

2

Solution :

tan θθθθ =( )

××××

5

25 9.8

2

tan θθθθ = 0.1021

∴∴∴∴ θ θ θ θ = tan–1 (0.021) = 50 50′′′′

∴∴∴∴ µ µ µ µ =v

rg

2

= 0.1021

MAHESH TUTORIALS SCIENCE .. 3

Circular Motion

Solution :

Since, v2

L= 5rg

∴∴∴∴ TL

= m5rg

+ gr

= 6mg

= 6 × × × × 1 × × × × 9.8 = 58.8 N∴∴∴∴ T

L= 58.8 N

Since, v2

M= 3rg

TM

= m3rg

r

= 3 mg

= 3 × × × × 1 × × × × 9.8 = 29.4 N∴∴∴∴ T

M= 29.4 N

Since, v2

H= rg

TH

= mrg

– gr

= 0

∴∴∴∴ TH

= 0

7. An object of mass 0.5 kg attached to arod of length 0.5 m is whirled in avertical circle at constant angular speed.If the maximum tension in the string is5 kg wt Calculate.i) Speed of stoneii) Maximum number of revolutions

it can complete in a minute.Given :

m = 0.5 kg,

r = l = 0.5 m,

Tmax

= 5 kg wt. = 5 ×××× 9.8 N

To Find :

i) speed of stone, v = ?

ii) maximum number of revolutions

in one minute = ?

Formula :

i) Tmax

=mv

r

2

+ mg

ii) n =v

2 rπ

Solution :

Tmax

=mv

r

2

+ mg

∴∴∴∴ v2 =r

m(T

max – mg)

∴∴∴∴ v2 = r maxT

– gm

= 0.55×9.8

– 9.80.5

= 49 – 4.9 = 44.1

∴∴∴∴ v = 44.1 = 6.64 m/s

nmax

=v

2 rπ[ ]∵v = rωωωω

=6.64

2 3.14 0.5× ×× ×× ×× ×

= 2.115 r.p.s

∴∴∴∴ nmax

= 2.115 × × × × 60n

max= 126.9 r.p.m

8. A motor van weighing 4400 kg roundsa level curve of radius 200 m onunbanked road at 60 km/hr. Whatshould be minimum value of coefficientof friction to prevent skidding ? At whatangle the road should be banked for thisvelocity ?

Given :m = 4400 kgr = 200 mv = 60 km/hr

= 60 × × × × 5

18=

50

3m/s

g = 9.8 m/s2

To Find :i) µµµµ = ?ii) θθθθ = ?

Formula :

i) v = rgµµµµ

ii) tan θ θ θ θ =v

rg

2

Solution :

µµµµ =v

rg

2

.. 4 MAHESH TUTORIALS SCIENCE

Circular Motion

10. A pilot of mass 50 kg in a jet aircraftwhile executing a loop–the–loop withconstant speed of 250 m/s. If the radiusof circle is 5 km, compute the forceexerted by seat on the piloti) at the top of loopii) at the bottom loop.

Given :m = 50 kgr = 5 kmv = 250 m/s = 5 × × × × 103 m

To Find :i) F

top = ?

ii) Fbottom

= ?

Formula :

i) Ftop

= N1 =

mv

r

2

– mg

ii) Fbottom

= N2 =

mv

r

2

+ mg

9. A string of length 0.5 m carries a bob ofmass 0.1 kg at its end. It is used as aconical pendulum with a period 1.41 sec.Calculate angle of inclination of stringwith vertical and tension in the string.

[Note : There is a correction in the question]Given :

l = 0.5 mm = 0.1 kgT = 1.41 sec

To Find :θθθθ = ?

tension in the string T′′′′ = ?

Formula :

T = 2ππππcos

g

l θθθθ

Tension, (T′′′′) = mg

cosθθθθ

Solution :

T = 2ππππcos

g

l θθθθ

∴∴∴∴ 1.41 = 2 × × × × 3.142 0.5 cos

9.8

× θ× θ× θ× θ

∴∴∴∴1.41

2 3.142××××=

cos

19.6

θθθθ

∴∴∴∴ 1.41

2 3.142

2

××××=

cos

19.6

θθθθ

∴∴∴∴ cos θθθθ = 0.9868∴∴∴∴ θ θ θ θ = cos–1(0.9868))))

∴∴∴∴ θ θ θ θ = 9019′′′′

=( )50 /3

200 9.8

2

××××=

25

18 9.8××××

∴∴∴∴ µµµµ = 0.1417

tan θθθθ = 0.1417

∴∴∴∴ θ θ θ θ = tan–1 (0.1417)∴∴∴∴ θ θ θ θ = 804′′′′

T′′′′ =mg

cosθθθθ

∴∴∴∴ T′′′′ =( )

0.1 9.8

cos 9 19'0

××××

=0.98

0.9868

∴∴∴∴ T′′′′ = 0.993 N

T′′′′ cos θθθθ

Tθθθθ

θθθθ

mg

Pilot

PilotN

2

N1

mg

mg

2mv

r

2mv

r

MAHESH TUTORIALS SCIENCE .. 5

Circular Motion

11. A ball is released from height h alongthe slope and move along a circular trackof radius R without falling verticallydownwards as shown in the figure.

Show that h = 5

2R

Proof : PE = KE

mgh =1mv

22B

∴∴∴∴ mgh =5

2 mgR

V = 5Rg

V = 5Rg

B

2B

Hence according to law of conservationof energy,

∴∴∴∴ gh =5

2 gR

∴∴∴∴ h =5

2 R

12. A block of mass 1 kg is released from Pon a frictionless truck which ends inquarter circular track of radius 2 m atthe bottom as shown in figure. What isthe magnitude of radial acceleration andtotal acceleration of the block when itarrive at Q ?

Given :H = 6mr = 2mu = 0 (... body starts from rest)

To Find :i) a

R= ?

ii) aTotal

= ?Formula :

i) aR

=v

r

2

ii) aTotal

= a + aR T2 2

Solution :Height lost by the body = 6 – 2

= 4mFrom equation of motionv2 = u2 + 2gh

∴∴∴∴ v2 = 0 + 2 ×××× 9.8 ×××× 4 = 78.4

aR

=v

r

2

=78.4

2

∴∴∴∴ aR

= 39.2 m/s2

aT

= g = 9.8 m/s2

aTotal

= ( ) ( )39.2 + 9.82 2

= 1536.64 + 96.04

= 1632.68

∴∴∴∴ aTotal

= 40.4 m/s2

P

H = 6m

2m

Q

Solution :

Ftop

=( )××××

××××

50 250

5 10

2

2 – 50 ×××× 9.8

= 625 – 490

∴∴∴∴ Ftop

= 135 N

Fbottom

=( )××××

××××

50 250

5 10

2

3 + 50 ×××× 9.8

= 625 + 490

∴∴∴∴ Fbottom

= 1115 N

R

B

h

.. 6 MAHESH TUTORIALS SCIENCE

Circular Motion

14. The length of hour hand of a wrist watchis 1.5 cm. Find magnitude ofi) angular velocityii) linear velocityiii) angular accelerationiv) radial accelerartionv) tangential accelerationvi) linear acceleration of a particle on

tip of hour hand.Given :

T = 12 ×××× 60 ×××× 60 = 43200 sr = 1.5 cm

= 1.5 ×××× 10–2 m

13. A circular race course track has a radiusof 500 m and is banked to 100. If thecoefficient of friction between tyres ofvehicle and the road suface is 0.25.Compute.i) the maximum speed to avoid

slippingii) the optimum speed to avoid wear

and tear of tyres (g = 9.8 m/s2)Given :

r = 500θθθθ = 100

µµµµ = 0.25

To Find :i) v

max= ?

ii) v0

= ?Formula :

i) vmax

=

+ tanrg

1+ tans

s

µ θµ θµ θµ θ

µ θµ θµ θµ θ

ii) v0

= rg tan θθθθ

Solution :

vmax

=

0.25+ tan10500 9.8

1 – 0.25 tan10

0

0××××

××××

∴∴∴∴ vmax

= 46.72 m/s

v0

= 500 9.8× tan100××××

∴∴∴∴ v0

= 500 9.8×0.176××××

∴∴∴∴ v0

= 29.37 m/s

To Find :i) ωωωω = ?ii) v = ?iii) αααα = ?iv) a

R= ?

v) aT

= ?vi) a = ?

Formula :

i) ωωωω =2

T

ππππ

ii) v = rωωωω

iii) αααα =d

dt

ωωωω

iv) aR

= vωωωωv) a

T= αααα r

vi) a = a a2 2R T+ a a2 2

R T+

Solution :

ωωωω =2

T

ππππ

ωωωω =2 3.142

43200

××××

∴∴∴∴ ωωωω = 1.454 ×××× 10–4 rad/s

v = rωωωω

v = 1.5 ×××× 10–2 ×××× 1.46 ×××× 10–4

∴∴∴∴ v = 2.19 ×××× 10–6 m/sSince angular velocity of hour hand isconstant, αααα = 0

∴∴∴∴ aT

= αααα raT

= 0aR

= vωωωωaR

= 2.182 ×××× 16–6 ×××× 1.454 ×××× 10–4

∴∴∴∴ aR

= 3.175 ×××× 10–10 m/s2

a = a a2 2R T+

= a 02R +

= aR

a = 3.175 ×××× 10–10

∴∴∴∴ a = 3.175 ×××× 10–10 m/s2