Circular Motion
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Angular Displacement, Angular Velocity, Angular Acceleration, Frequency and Time Period
Angular Displacement
Angular Velocity (ω)
Its SI unit is rad s−1.
Angular Acceleration
Its SI unit is rad s−2.
Frequency
Its SI unit is hertz (Hz).
Time Period (T)
Since, the angular displacement is 2π in one revolution,
ω = 2πT
Since, f = 1T
∴ ω = 2πfRelation Between Linear Velocity and Angular Velocity
Let us consider a body moving in a circular path of radius r. Suppose the body is initially at point A and after time t it reaches point B with angular displacement θ and linear displacement s.
From figure:
s = rθ …(i)
Differentiating equation (i) with respect to time:
dsdt = r dθdt
Since, dsdt = v and dθdt = ω
∴ v = rω …(ii)Again, differentiating equation (ii) with respect to time:
dvdt = r dωdt
Since, dvdt = a and dωdt = α
∴ a = αr …(iii)Expression for Centripetal Acceleration
Let us consider a body of mass m moving in a circular path of radius r with uniform angular velocity ω. Suppose the body is initially at point A and after time t it reaches point P(x, y) with angular displacement θ = ωt.
Position vector after time t:
r⃗ = îx + ĵy
= î(r cosθ) + ĵ(r sinθ)
= r[î cosωt + ĵ sinωt] …(i)
Velocity at time t:
v⃗ = dr⃗dt
= ddt[r(î cosωt + ĵ sinωt)]
v⃗ = ωr[−î sinωt + ĵ cosωt] …(ii)Acceleration at time t:
a⃗ = dv⃗dt
= ω²r[−î cosωt − ĵ sinωt]
= −ω²r(î cosωt + ĵ sinωt)
Using equation (i):
a⃗ = −ω²r⃗ …(iii)The negative sign shows that acceleration is directed towards the centre of the circular path.
This is the required expression for centripetal acceleration.
Expression for Centripetal Force
Centripetal force = mass × centripetal acceleration
F = m × ω²r
Since, v = ωr,
F = m(vr)²r
∴ F = mv²r …(v)Motion of Cyclist in Circular Path
Let us consider a cyclist moving in a circular path of radius r with uniform velocity v. Let R be the reaction and θ be the angle of inclination with vertical. The reaction R can be resolved into two components.
Horizontal component provides the necessary centripetal force:
R sinθ = mv²r …(i)
Vertical component balances the weight of cyclist:
R cosθ = mg …(ii)
Dividing (i) by (ii):
tanθ = v²rg …(iii)The source notes that θ increases when v is increased and r is decreased.
Motion of Car in Curved Circular Path
Let us consider a car (vehicle) of mass m moving in a circular path of radius r with constant velocity v. Let R1 and R2 be the reactions at the left and right side of tires, so total reaction R = R1 + R2.
Total reaction balances the weight:
R = mg …(i)
Frictional force provides necessary centripetal force:
Ff = Fc
μR = mv²r
Using R = mg:
μmg = mv²r
∴ v = √(μrg)This is the maximum velocity with which a vehicle can take a safe circular turn of radius r.
Banking of Road
Let us consider a car of mass m moving on a banked road with uniform velocity v. Let θ be the angle of banking and R be the total reaction of the car. Then R can be resolved into two components.
Horizontal component provides necessary centripetal force:
R sinθ = mv²r …(i)
Vertical component balances the weight:
R cosθ = mg …(ii)
Dividing (i) by (ii):
tanθ = v²rgThis is the required expression for banking of road.
Motion in a Vertical Circle
At point P:
Fc = T − mg cosθ
T = Fc + mg cosθ
T = mv²r + mg cosθAt point A, θ = 0°
Tmax = mv²r + mg
At point C, θ = 180°
Tmin = mv²r − mg
At points B and D, θ = 90° or 270°
T = mv²r
Numericals: Vertical Circle and Banking of Road
L.B. 1(A)
An object of mass 8.0 kg is whirled round in a vertical circle of radius 2 m with a constant speed of 6 m s−1. Calculate the maximum and minimum tensions in string.
m = 8 kg, r = 2 m, v = 6 m s−1.
Tmax = mv²r + mg
= 8(6)²2 + 8(10)
Tmax = 224 N
Tmin = mv²r − mg
= 8(6)²2 − 8(10)
Tmin = 64 N
L.B. 1(B)
A mass of 0.2 kg is rotated by a string at a constant speed in a vertical circle of radius 1 m. If the minimum tension in the string is 3 N, calculate the magnitude of the speed and the maximum tension in the string.
m = 0.2 kg, r = 1 m, Tmin = 3 N.
Tmin = mv²r − mg
3 = 0.2v²1 − 0.2(10)
v = 5 m s−1
Tmax = mv²r + mg
= 0.2(5)²1 + 0.2(10)
Tmax = 7 N
B.1(C)
At what angle should a circular road be banked so that a car running at 50 km hr−1 is safe to go round the circular turn of 200 m radius?
v = 50 km hr−1 = 13.89 m s−1, r = 200 m.
tanθ = v²rg
tanθ = (13.89)²200(10) ≈ 0.096
θ = tan−1(0.096)
θ = 5.5°
B.1(D)
An object of mass 4.0 kg is rotated in a vertical circle of radius 1 m with a constant speed of 3 m s−1. Calculate the maximum tension in the string.
m = 4 kg, r = 1 m, v = 3 m s−1.
Tmax = mv²r + mg
= 4(3)²1 + 4(10)
= 36 + 40
Tmax = 76 N
Coin on a Rotating Disc
B.1(E)
A coin placed on a disc rotates with speed of 33⅓ rev min−1, provided that the coin is not more than 10 cm from the axis. Calculate the coefficient of static friction between the coin and the disc.
Revolution per minute = 33⅓ rev min−1 = 1003 rev min−1.
Frequency per second = 1003 × 160 = 59 rev s−1.
r = 10 cm = 0.1 m.
ω = 2πf = 2π × 59 = 10π9 rad s−1.
v = rω = 0.1 × 10π9 = π9 m s−1.
Using v = √(μrg):
μ = v²rg
μ = (π/9)²0.1(10) = 0.121945
Coefficient of static friction ≈ 0.122
Conical Pendulum (Horizontal Pendulum)
Let us consider a small bob of mass m suspended by a string of length l from a rigid support. The bob is whirled in a horizontal circle of radius r with constant velocity v, so that the string is inclined by angle θ with vertical and the vertical height is h.
T sinθ = mv²r …(i)
T cosθ = mg …(ii)
Dividing (i) by (ii):
tanθ = v²rg …(iii)
Also from figure, tanθ = rh …(iv)
From (iii) and (iv):
rh = v²rg
rv = √(hg) …(v)If t is the time period of conical pendulum:
t = 2πrv …(vi)
Using equation (v) in (vi):
t = 2π√(hg) …(vii)Also from figure, cosθ = hl
h = l cosθ …(viii)
Using (viii) in (vii):
t = 2π√(l cosθg)Conical Pendulum Numericals
B.2(A)
A bob of mass 200 gram is whirled in a horizontal circle of radius 50 cm by a string inclined at 30° to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle.
m = 200 g = 0.2 kg, r = 50 cm = 0.5 m, θ = 30°.
T cosθ = mg
T cos30° = 0.2(10)
T = 2.3 N
And, T sinθ = mv²r
2.3 sin30° = 0.2v²0.5
v = 1.699 m s−1
B.2(B)
An object of mass 0.5 kg is rotated in a horizontal circle by a string 1 m long. The maximum tension in the string before it breaks is 50 N. What is the greatest number of revolutions per second of the object?
m = 0.5 kg, l = 1 m, T = 50 N.
T cosθ = mg
50 cosθ = 0.5(10)
cosθ = 0.1
Time period:
t = 2π√(l cosθg)
= 2(3.14)√(1(0.1)10) = 0.628 s
f = 1t = 10.628
f = 1.59 rev s−1
B.2(C)
A certain string breaks when a weight of 25 N acts on it. A mass of 500 gram is attached to one end of the string of 1 m long and is rotated in a horizontal circle. Find the greatest number of revolutions per minute which can be made without breaking the string.
T = 25 N, m = 500 g = 0.5 kg, l = 1 m.
T cosθ = mg
25 cosθ = 0.5(10)
cosθ = 525 = 15
t = 2π√(l cosθg)
= 2 × 227 √(1 × (1/5)10)
t = 0.8889 s
Frequency per second = 10.8889 = 1.1249 rev s−1.
Revolution per minute = 1.1249 × 60
= 67.49 rev min−1
B.2(D)
A stone with mass 0.8 kg is attached to one end of a string 0.9 m long. The string will break if its tension exceeds 600 N. The stone is whirled in a horizontal circle, the other end of the string remains fixed. Find the maximum speed the stone can attain without breaking the string.
m = 0.8 kg, l = 0.9 m, T = 600 N and, as used in the source, r = l = 0.9 m.
T cosθ = mg
600 cosθ = 0.8(10)
θ = cos−1(8/600) = 89.23°
Also, T sinθ = mv²r
600 sin89.23° = 0.8v²0.9
v = 25.98 m s−1
B.2(E)
A mass of 1 kg is attached to the lower end of a string 1 m long whose upper end is fixed. The mass is made to rotate in a horizontal circle of radius 60 cm. If the circular speed of the mass is constant, find the tension in the string and the period of motion.
m = 1 kg, l = 1 m, r = 60 cm = 0.6 m.
From figure, sinθ = rl = 0.61
θ = 36.86°
T cosθ = mg
T = 10cos36.86° = 100.8
T = 12.5 N
t = 2π√(l cosθg)
= 2 × 227 × √(1 × 0.810)
t = 1.77 s
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