CIRCULAR MOTION
EXERCISES, Questions 11 to 20
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?
ANSWER: Max Angular Speed of the fan
ω = 1500 rpm = 1500x2π/60 rad/s =50π rad/s
Radius r =120/2 cm =60 cm =0.60 m
Mass of the particle =1 g = 0.001 kg
Force experienced by the particle = mω²r
At full speed, this force = 0.001x(50π)²x0.60 N
=0.001x2500x0.60 π² =2.5x0.60π² =1.5π² N =1.5x3.14x3.14 N
= 14.8 N
Since the particle is in a circular motion, it has a radial / centripetal acceleration (towards the center). So the force on the particle is also towards the center. Due to the inertia of the particle, it tends to move along the tangential direction with a uniform speed. Since it is stuck to the blade, the blade of the fan exerts a centripetal force on the particle and forces it to move in a circle.
It is the inertia of the particle that exerts the same magnitude of the force on the blade, i.e. 14.8 N is exerted by the particle on the blade along its surface.
12. A mosquito is sitting on an LP record disc rotating on a turntable at 33⅓ revolutions per minutes. The distance of the mosquito from the center of the turntable is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π²/81. Take g= 10 m/s².
ANSWER: Angular speed of the mosquito, ω=33⅓ rpm =33⅓x2π/60 =200π/180 =10π/9 rad/s.
Radius, r=10 cm =0.10 m.
Outward force on the mosquito =mω²r (where 'm' is mass of the mosquito).
=m.(100π²/81)x0.10
=10π²m/81
Weight of mosquito = mg = 10m N =Normal force
So Limiting frictional force =µ.10m N
Since the mosquito is sitting on the record, this limiting Frictional force > centrifugal force on the mosquito.
→ µ.10m>10π²m/81
→ µ>π²/81.
Hence proved.
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.
ANSWER: Let the mass of the bob be =m
Speed of the car =36km/h =36000/3600 m/s
= 10 m/s
The radius of the turn, r =10 m
The horizontal force (Outward)on the bob at the turn = mv²/r
=m.100/10 =10m N
Vertical force (Downward) on the bob
=Weight of bob, mg =10m N
Resultant of these two forces is balanced by the tension in the string. So the resultant and the string is in the same line. See the diagram below,
Let the string make an angle θ with the vertical. So the resultant of these two forces also make an angle θ with the downward force.
Tanθ=10m/10m =1
→ Ө=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 the instant.
ANSWER: Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.
i.e. T =mg+mv²/r
Here m =100 g =0.10 kg, v=1.4 m/s, r=1 m
So, T=0.10x10 + 0.10x1.4²/1 N
=1+0.196 N =1.196 N ≈ 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 the instant. You can use cosθ≈1-θ²/2 and sinθ≈θ for small θ.
ANSWER: Here θ=0.20 radian. Radial outward force on the bob due to circular motion =mv²/r = 0.10(1.4)²/1 =0.196 N.
Component of weight along the string line = mg.cosθ
=0.10x9.8 (1-θ²/2) = 0.98(1-θ²/2) =0.98(1-0.04/2) =0.98(1-0.02) =0.96 N
Total outward force on the bob along the string line = 0.196+0.96 =1.156 N ≈ 1.16 N
This outward force on the string is balanced by the tension in the string which is also equal to 1.16 N.
16. Suppose the amplitude of a simple pendulum having a bob of mass m is θ. Find the tension in the string when the bob is at its extreme position.
ANSWER: Tension T in the string is equal to the total radial force on the bob at the instance. So T=mv²/r+mg.cosθ, (See figure below),
 |
| Diagram for Problem no 16 |
where v is the speed of the bob. The first term is the force due to circular motion while the second term is the component of the weight along the radial line. In the given problem the bob is at its extreme position so v=0.
→T=mg.cosθ
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 the earth 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?
ANSWER: (a) Let time period of the earth for one rotation around its own axis, T = 24 hrs.
The angular speed of the earth ω=2π/T =2π/(24x3600) radians/s
The outward force due to it=mω²R (R-radius of the earth, taking it=6400 km)
It is this outward force by which the true weight of the person reduces.
The fraction by which the weight reduces = mω²R/mg
=ω²R/g =4π²x6400x1000/(24x24x3600x3600x9.8)
=80π² /(6x3x36x36x9.8) =80x9.86/228614.4 = 0.00345 =3.45x10-3
(b)In this case the outward force due to the rotation of earth= Half the true weight
mω²R=½mg
→2(2π/T)²R=g
→8π²R=gT²
→T²=8π²R/g =8x9.86x6400x1000 /9.8 =51513469.4
→T=7177 s =7177/3600 hr =1.994 hr ≈ 2 hrs
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 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?
ANSWER: If the vehicle is going at a speed of 36 km/h the equilibrium of forces is as in figure below,
So, N.cosθ=mg and N.sinθ=mv²/r
Dividing later by former, we get tanθ=v²/gr
Here r=20 m, v=36000/3600 m/s =10 m/s
So, tanθ=100/10x20 =1/2
(Taking g=10 m/s²)
But it is the case where friction does not come into the picture.
Let the maximum velocity without skidding up be V. Now a frictional force, F will act on the vehicle along the plane as shown in the picture below.
F=µN
Now, N.sinθ + F.cosθ = mV²/r (Horizontal equilibrium of forces)
→N.sinθ = mV²/r-F.cosθ ..... (i)
N.cosθ-F.sinθ = mg (Vertical equilibrium of forces) →N.cosθ = mg+F.sinθ ..... (ii)
Dividing (i) by (ii) we get,
tanθ = (mV²/r-F.cosθ)/(mg+F.sinθ) =1/2
(we got tanθ=1/2, so sinθ=tanθ/√(1+tan²θ)=1/√5
and cosθ=1/√(1+tan²θ)=2/√5)
→ mg+F.sinθ = 2mV²/r-2F.cosθ
→ F(sinθ+2cosθ) = m(2V²/r-g)
→ µN(1/√5 + 2x2/√5) =m(2V²/20 - 10)
→ 0.4 N(5/√5)= m(V²-100)/10
→4√5 N=m(V²-100) ..... (iii)
Resolving forces perpendicular to surface, we get
mg.cosθ +(mV²/r)sinθ=N
→N = 10m. 2/√5 + mV²/20√5 = m(20/√5+V²/20√5)
→N =m(400+V²)/20√5
→20√5N = m(400+V²) ..... (iv)
Dividing (iii) b(iv) we get,
1/5= (V²-100)/(V²+400)
→V²+400=5V²-500
→4V² = 900
→V² =900/4 →V = 30/2 =15 m/s = 15x3600/1000 =54 km/hr
Now let us consider the case for minimum velocity 'u' on the turn so that vehicle does not slip down.
In this case, the force of friction will act upward along the plane. See the picture below:-
N.sinθ=F.cosθ+mu²/r -------- (v)
(Horizontal equilibrium of forces)
(Taking g=10 m/s²)
But it is the case where friction does not come into the picture.
Let the maximum velocity without skidding up be V. Now a frictional force, F will act on the vehicle along the plane as shown in the picture below.
 |
| Forces for maximum velocity V |
F=µN
Now, N.sinθ + F.cosθ = mV²/r (Horizontal equilibrium of forces)
→N.sinθ = mV²/r-F.cosθ ..... (i)
N.cosθ-F.sinθ = mg (Vertical equilibrium of forces) →N.cosθ = mg+F.sinθ ..... (ii)
Dividing (i) by (ii) we get,
tanθ = (mV²/r-F.cosθ)/(mg+F.sinθ) =1/2
(we got tanθ=1/2, so sinθ=tanθ/√(1+tan²θ)=1/√5
and cosθ=1/√(1+tan²θ)=2/√5)
→ mg+F.sinθ = 2mV²/r-2F.cosθ
→ F(sinθ+2cosθ) = m(2V²/r-g)
→ µN(1/√5 + 2x2/√5) =m(2V²/20 - 10)
→ 0.4 N(5/√5)= m(V²-100)/10
→4√5 N=m(V²-100) ..... (iii)
Resolving forces perpendicular to surface, we get
mg.cosθ +(mV²/r)sinθ=N
→N = 10m. 2/√5 + mV²/20√5 = m(20/√5+V²/20√5)
→N =m(400+V²)/20√5
→20√5N = m(400+V²) ..... (iv)
Dividing (iii) b(iv) we get,
1/5= (V²-100)/(V²+400)
→V²+400=5V²-500
→4V² = 900
→V² =900/4 →V = 30/2 =15 m/s = 15x3600/1000 =54 km/hr
Now let us consider the case for minimum velocity 'u' on the turn so that vehicle does not slip down.
In this case, the force of friction will act upward along the plane. See the picture below:-
 |
| Forces for minimum velocity u |
N.sinθ=F.cosθ+mu²/r -------- (v)
(Horizontal equilibrium of forces)
N.cosθ+F.sinθ=mg
(Vertical equilibrium of forces)
→N.cosθ = mg-F.sinθ -------- (vi)
Dividing (v) by (vi) we get,
tanθ=(F.cosθ+mu²/r)/(mg-F.sinθ) =1/2 (Since tanθ=1/2)
→ (mg-F.sinθ) =2Fcosθ+2mu²/r
→F(sinθ+2cosθ)=m(g-2u²/r)
→µN(1/√5+2*2/√5)=m(10-2u²/20)
→0.4N√5=m(100-u²)/10
→4√5N=m(100-u²) ---- (vii)
Now resolving the forces along N,
N=mg.cosθ+mu²/r*sinθ =m(gcosθ+u²/r*sinθ)
→N=m(10*2/√5+u²/20√5)
→N=m(u²+400)/20√5
→20√5N=m(u²+400) ---(viii)
Dividing (vii) by (viii), we get
1/5=(100-u²)/(u²+400)
→u²+400=500-5u²
→6u² = 100
→u²=100/6
→u=10/√6 m/s =10*3600/√6*1000 km/hr
=36/√6 km/hr = 14.7 km/hr
(Vertical equilibrium of forces)
→N.cosθ = mg-F.sinθ -------- (vi)
Dividing (v) by (vi) we get,
tanθ=(F.cosθ+mu²/r)/(mg-F.sinθ) =1/2 (Since tanθ=1/2)
→ (mg-F.sinθ) =2Fcosθ+2mu²/r
→F(sinθ+2cosθ)=m(g-2u²/r)
→µN(1/√5+2*2/√5)=m(10-2u²/20)
→0.4N√5=m(100-u²)/10
→4√5N=m(100-u²) ---- (vii)
Now resolving the forces along N,
N=mg.cosθ+mu²/r*sinθ =m(gcosθ+u²/r*sinθ)
→N=m(10*2/√5+u²/20√5)
→N=m(u²+400)/20√5
→20√5N=m(u²+400) ---(viii)
Dividing (vii) by (viii), we get
1/5=(100-u²)/(u²+400)
→u²+400=500-5u²
→6u² = 100
→u²=100/6
→u=10/√6 m/s =10*3600/√6*1000 km/hr
=36/√6 km/hr = 14.7 km/hr
19. A motorcycle has to move with a constant speed on an over-bridge which is in the form of the circular arc of radius R and has a total length L. (a) Suppose the motorcycle starts from the highest point. 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?
ANSWER: Due to circular motion, the motorcycle will experience a radial force = mv²/R, where m is mass of motorcycle and v is the velocity.
(a) At the highest point, this force is upwards while the weight of the motorcycle is downward. So not to break contact with the road the maximum velocity should be such that
mv²/R=mg → v²=Rg →v=√(Rg)
(b) Now v=1/√2 *√(Rg) =√(Rg/2)
Let at an angular distance θ from the vertical it is about to lose the contact. It is clear that the contact-making force i.e. weight 'mg' is now not in line with the contact breaking force mv²/R. Only a component of mg will counter it. Equilibrium of the forces in the radial direction is,
Picture for Q 19
mv²/R =mg.cosθ
→v²=Rg.cosθ
→(√(Rg/2)² =Rg.cosθ
→Rg/2=Rg.cosθ
→cosθ=1/2
→θ=60° =π/3 Radian
This point will be at a distance l=Rθ from the highest point along the bridge.
i.e. l=Rπ/3
(c) As we have seen above in (b) that v²=Rg.cosθ
In this problem, Rg is constant, so as θ increases cosθ decreases and so is 'v'. It means at the start or end of the overbridge where θ is maximum, the motorcycle will first lose contact.
Distance along the bridge from the top at this point =L/2
so θ=(L/2)/R=L/2R
Putting this value in v²=Rg.cosθ
v²=Rg.cos(L/2R)
→v=√{Rg.cos(L/2R)}
20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt=a. The friction coefficient between the road and the tyre is µ. Find the speed at which the car will skid.
ANSWER: Since the circular motion is non-uniform, the car will have two-component of acceleration, Radial and tangential.
Magnitude of Radial acceleration =v²/R .
Magnitude of Tangential acceleration =dv/dt =a
Magnitude of total acceleration =√{(v²/R)²+a²}
Horizontal Force on the car, P = m√{(v²/R)²+a²}
Normal Force on the car, N= mg
Limiting Frictional force on the car, F =µN =µmg
Just when the car is about to skid, P=F
→m√{(v²/R)²+a²}=µmg
→ √{(v²/R)²+a²}=µg
→ (v²/R)²+a²=µ²g²
→ (v²/R)²=µ²g²-a²
→ v⁴/R² = µ²g²-a²
→ v⁴ = (µ²g²-a²)R²
→v = {(µ²g²-a²)R²}¹/⁴
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CHAPTER- 20 - Dispersion and Spectra
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CHAPTER- 20 - Dispersion and Spectra
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