Circular motion/ machines/calculation on circular motion/Questions and Answers

1. Find the acceleration of the moon with respect to the earth from the following data. Distance between the earth and the moon = 3.85 × 10 ³ km and the time taken by the moon to complete one revolution around the earth = 27.3 days

Solution

Distance between Earth and moon 

r=3.85x10⁵km = 3.85x10⁸m

T=27.3days=24x3600 x (27.3)sec = 2.36 x 10⁶sec

v = 2πr/T = 2x3.14 x3.85x10⁸/2.36 x 10⁶ =1035.42m/sec

a= v²/r = (1024.42)²/3.85x10⁸

⁼ 0.00273msec⁻² = 2.73 x 10⁻³m/sec⁻²

2.Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth 12800 km and it takes 24 hours for the earth to complete one revolution about its axis. 

SOLUTION

Diameter of earth = 12800km

Radius R= 6400km = 64 x 10⁵m

a= 2πR/T = 2x3.14 x 64 x 10⁵/  24 x3600 msec⁻¹

=465.185msec⁻¹

a V²/R = (46.5185)/64x10⁵ =0.0338m/sec²

A particle moves in a circle of radius 10 cm at a speed given by v 2.0 t where u is in cm/s and t in seconds. = (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s. 

SOLUTION 3

V = 2t, r = 1cm 

a) Radial acceleration at t = 1 sec. a = v²/r= 2² /=  4cm/sec² 

b) Tangential acceleration at t = 1 sec.

    a=dv/dt= d(2t)/dt  = 2cm/sec²  

c)Magnitude of acceleration at t = 1se

a = √4² +2² = √20 cm/sec²

4. A scooter weighing 150 kg together with its rider moving at 36 km/hrs is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

SOLUTION

Given that m= 150kg, v= 36km/hr = 10m/sec, 

r =30 

Horizontal force needed is mv²/r= 150 × (10)²/ 30 = 150 x 100/30 =500N

 5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking ? 

SOLUTION

R cos θ = mg......i 

 R sin θ = mv²/ r ..(0) ..(ii)  Dividing equation (i) with equation (ii) 

Tan θ= mv² /rmg  =v²/rg 

 v = 36km/hr = 10m/sec,

r = 30m

 Tan θ= v² /rg=100/30 x 10

Tanθ =(1/3)

θ=tan⁻¹ (1/3)

6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking? 

SOLUTION

Radius of Park = 

r = 10m 

speed of vehicle = 18km/hr = 5 m/sec 

Angle of banking 

tanθ = ν²/rg  ⇒ 

θ=  tan⁻¹ ( 25/100 )

 =

tan⁻¹(1/4)

7. If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at 18 km/hr does not skid? 

SOLUTION

The road is horizontal (no banking)

 mv² /R = μN and N = mg So mv²/R = μ mg 

v = 5m/sec, 

R= 10m

  = > 25/10 >> = μg = μ =25 /100 = 0.25

8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

SOLUTION

Angle of banking θ=30° Radius r= 50m 

tan θ=  v²/rg 

tanθ =30º = v²/rg

 ==> 1 /√3 = v²/rg

V²= rg/√3

=  50x10/ √3 

V= √(3 500/√3)

v= 17m/sec.

9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron the proton in a circle of radius round the goes round - 11 5.3 x 107 Find the speed of the electron in 31 10-³¹ kg the ground state. Mass of the electron = 9.1 x 10 and charge of the electron = 1.6 × 10-¹⁹ C. 

SOLUTION

Electron revolves around the proton in a circle having proton at the centre. Centripetal force is provided by coulomb attraction 

r= 5.3 X 10⁻¹¹ m 

m = mass of electron =9.1 * 10⁻³kg charge of electron 1.6x 10⁻¹⁹ c.

 mv²/r =kq²/r²

v² =kq²/rm

 9x10⁹ x 1.6 x 1.6x10⁻³⁸/ 5.3x10⁻¹¹9.1x10⁻³¹

 =23.04 /48.23*10¹³ v²=0·477x10²³ = 4.7-10¹² ⇒v= √4.7×10¹²=2.2 x 10⁶m/sec

10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stone can have at the highest point of the circle. 

SOLUTION

At the highest point of a vertical circle 

mv² /R = mg ⇒V²=Rg

v= √Rg

11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface ? 

Solution

A ceiling fan has a diameter 120cm. Radius=r=60cm =0/6m Mass of particle on the outer end of a brade is 1g.

n= 1500 rev/min =25 rev/sec 

ω=2πn=2π x 25=157.14 Force of the particle on the blade =Mrω² = (0.001) 0.6 x (157.14) =14.8N

 The fan runs at a full speed in circular path. This exerts the force on the particle (inertia) The particle also exerts a force of 14.8N on the blade along its surface.

12. A mosquito is sitting on an L.P. record disc rotating on a turn table at 33- revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than ¹/81. Take g= 10 m/s ². 

 SOLUTION

A mosquito to is sitting on an L.P. record disc & rotating on a turn table at n = 1/3 x33 rpm

 = 100/ 3x60rps 

ω=2πn = 2π x 100/180

ω =10π/9 radsec⁻¹

r =10cm = 0.1m

g = 10ms⁻²

μmg≥mrω²

μ = rω²/g

μ≥ 0.1x (10π/9)²/10

μ≥π²/81

13. A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s².

A pendulum is suspended from the ceiling of a car taking a turn 

r = 10m, 

v = 36km/hr = 10 m/sec, g=10m/sec² 

T sin θ = mv²/r.........i

 T cos θ= mg......i

 sin/cos = mv²/rmg 

⇒tan θ= v²/rg

θ=tan⁻¹(v²/rg)

tan⁻¹(100 /10x10) 

tan⁻¹ (1)= 45°

14. The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1'4 m/s at the lowest point in its path. Find the tension in the string at this instant.

SOLUTION

At the lowest pt.

 T = mg + mv²/r 

Here m= 100g = 1/10 kg,

r = 1m 

v = 1.4 m/sec 

T = mg+mv²/r =1/10 x 

9.8 x (1.4/)²/10 = 0.98 + 0.196=1.176 =1.2N

mv² 1 x9.8x = 0.98 +0.196 = 1.176 = 1.2 N

 15. Suppose the bob of the previous problem has a speed of 1-4 m/s when the string makes an angle of 0-20 radian with the vertical. Find the tension at this instant. You can use cos=1-0/2 and sine for small 0. 

SOLUTION

Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian. 

m = 100g = 0.1kg, 

 r = 1m, v = 1.4m/sec. T-mg cos θ= mv² /R

 ⇒T= mv² /R + mg cos θ

⇒T= 0.1x (1.4)² /1 +(0.1)x9.8 x (1- θ²/2)

⇒T = 0.196 +9.8 

x (1 -(2)²/2  )

(.. cos θ= 1 -θ²/2for small θ) 

 ⇒ T = 0.196+ (0.98) x (0.98) = 0.196 +0.964 = 1.156N = 1.16 N

16. Suppose the amplitude of a simple pendulum having a bob of mass m is 0,. Find the tension in the string when the bob is at its extreme position.

SOLUTION

At the extreme position, velocity of the pendulum is zero. So there is no centrifugal force. 

So T = mg cos θ,

 17. A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

SOLUTION

a) Net force on the spring balance. R = mg - mω²r 

So, fraction less than the true weight (3mg) is

= mg - (mg-mω²r)/mg

=ω²/g= (2π/24x3600)² x 6400 x 10³/10 =3.5 x 10⁻³

b)  when the balance reading is half the true weight..

= mg - (mg-mω²r)/mg = 1/2

ω²r = g/2 

ω =√g/2r = √10/2x6400 x10³ radsec⁻¹ 

Duration of the day is

T=2π/ω = 2πx√2x6400x10³/9.8 sec 

=2πx800/7x3600hr = 2hr

 18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 04, what are the possible speeds of a vehicle so that it neither slips down nor skids up? 

SOLUTION

Given, v = 36km/hr =10m/s. 

r = 20m, 

μ=0.4 

The road is banked with an angle, θ= tan⁻¹(v²/rg ) = tan⁻¹ 100/20 x 10

= tan ⁻¹(1/2) 

When the car travels at max. speed so that it slips upward, μR1, acts downward 

So, R₁ -mg cos θ - mv²₁/rsin θ = 0 .......i

 And uR₁+ mg sin 0 -mv²₁/r cos 0 = 0 ....  ii Solving the equation we get 

V₁=√ rg (tanθ-μ /1+μtanθ) 

√ 0.1/1.2 x 20x10

 = 4.082 m/s = 14.7 km/hr So, the possible speeds are between 14.7 km/hr and 54km/hr

19. A motorcycle has to move with a constant speed on an overbridged which is in the form of a circular are of radius. R and has a total length L. Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge ? 

SOLUTION

R = radius of the bridge.

 L total length of the over bridge 

a) At the highest pt.  mg =   mv²/R ⇒v²=Rg⇒v = √Rg 

b) Given, v = 1/   √2 x √Rg  suppose it loses contact at B. So, at B, mg cosθ= mv²/ R ⇒ v²=Rg cos θ = (√ Rv/2)²  =Rg cos θ=>  cos θ= 1/2= 60° = π/3 

θ =ι/r

ι  =θr= πR/3 = >

 So, it will lose contact at distance  πR/3   from highest point 

 c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end of the bridge for which α =L/2R So, mv²/ R = mg cos  α⇒ v=√  gR cos( L /2R ).

20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate da. The friction coefficient between the road and the tyre is u. Find the speed at which the car will skid. 

21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration a, at what angular speed will the block slip?

 Solution

a) When the ruler makes uniform circular motion in the horizontal plane

μmg = mω²L

ω = √μg/L 

b) when the ruler make uniformly accelerated circular motion

μmg =√mω²₂)²  + (mLα)² 

ω⁴₂ + α² =μ²g²/L²

ω₂=((μg/L)²- α²)¼

 22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of km/h on the track. 18 km/h (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed.g=10ms⁻²

Solution

Radius of the curves = 100m 

Weight = 100kg

 Velocity = 18km/hr = 5m/sec 

a) at B mg -mv²/R =N

===> N = (100×10) 100x25 /100 =1000-25= 975N 

At  d, N= mg +mv²/R = 1000 + 25 = 1025 N 

 b) At B & D the cycle has no tendency to slide. So at B & D. frictional force is zero. At 'C', mg sin = F => F = 1000 = 707N

 c) (i) Before 'C' mg cosθ- N = mv² /R ⇒N=mv²/R + mg cosθ  = 707 - 25=683N (ii) N-mg cos θ = mv²/ R

N=mv²/R

 N=mgcosθ+mg cosθ =25 +707 = 732N 

d) To find out the minimum desired coeff of friction, we have to consider a point just before C (where N is minimum) Now,

 μN= mg sinθ 

μ=682= 707 So. 

μ=1.037

23. In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod .Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 30 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted. by the rod on one of the kids.

SOLUTION

d=3m=> R=1.5m 

R= distance from the centre to one of the kids N = 20 rev per min = 20/60= 1/3 rev per/ sec ω=2πr= 2π/3 

m = 15kg

 Frictional force F = mrω² = 15 x (15)x (2π)²/9 = 5x (0.5) x 4π² = 10π²  Frictional force on one of the kids is 10π²

24.  A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle 8 with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is u. Find the range of the angular speed for which the block will not slip. 

Solution

if the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downward. Here, r = R sin 0 

R₁ - mg cos 0 -mω²,      (R sin 0) sin=0 

because r = R sin

and uR₁ mg sin-mω²₁ (R sin 0) cos 0 = 0 (0)  Substituting the value of R₁  it can be found out

  ω₁= (g(sinθ+μcos)/ Rsin(cos-μsinθ) )½

25. A particle is projected with a speed u at an angle 0 with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular are. What is the radius of this circle? This radius is called the radius of curvature of the curve at the point.

SOLUTION

Particle is projected with speed 'u' at an angle θ. At the highest pt. the vertical component of velocity is '0'

So, at that point, velocity = ucosθ

centripetal force = mu² cos²(θ/r)

At highest pt

mg=mv²/ r

r=u²cos²θ/g

 26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle 0/2 with the horizontal? 

27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is u. The block is given an initial speed . As a function of the speed u write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration

(dv/dt= Vᵈᵛ/dt) to obtain speed of the block after revolution 

28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity on in a circular path of radius R . A smooth groove AB of length L(<<R) is made on the surface of the table. The groove makes an angle 0 with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.

29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road  A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is μ = 0:58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate. 

30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R .A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

SOLUTIONS TO CONCEPTS CHAPTER 7 

1. Distance between Earth & Moon r=3.85 10 km= 3.85-10m

 T = 27.3 days = 24 3600 (27.3) sec=2.36 10° sec 22-3.14-3.85-10²-1025.42m/sec T 2.36 106 (1025 42² 3.85-10 -0.00273m/sec²=2.73 10 m/sec 2 Diameter of earth - 12800km Radius R-6400km 64 x 10 m v2R 2.  3.14x64x10 V= T 24x3600 a misec 465.185 (46.5185)-0.0338m/sec 64-10 R V² 3 V=2t r= 1cm a) Radia acceleration at t 1 sec. 1 -4cm/sec² b) Tangential acceleration at t- 1sec adv (21)=2cm/sec² d =(21)=2cm/sec² c) Magnitude of acceleration at t = 1sec a-√4² +2²= √20 cm/sec²

 4. Given that m= 150kg. v=36km/hr = 10m/sec. r = 30m 150 (10) 150-100-500N Horizontal force needed is my 150 (10) 30 30 5 in the diagram R cos 0 mg 40 Rsina my² Dividing equation () with equation (0) Tane my v² rmg rg y=36km/hr 10m/sec. r = 30m Tan 6 = 100 30x10 (1/3) 0-tan (1/3). Park=r=10m

 6. Radius of Park =r=10m speed of vehicle 18km/hr- 5 m/sec Angle of banking tani) a 25 100tan 0-tantan 100 -tan (1/4) circular motion

Distance between Earth & Moon r=3.85 10 km-3.85-10m T = 27.3 days = 24 3600 (273) sec=2.36 10° sec