CBSE Class 9 Science Advanced Physics Circular Motion Short Answer Questions Set 01

Short Answer Type

Question 1. Two particles of equal mass are revolving in circular orbits of radii r₁ and r₂ with the same period. Show that v₁ : v₂ = r₁ : r₂.
Answer:
The period T is the same, so the angular velocity ω = 2π/T is the same for both.
Linear speed v = rω, so v₁ = r₁ω and v₂ = r₂ω.
Dividing, v₁/v₂ = r₁/r₂, that is v₁ : v₂ = r₁ : r₂.
In simple words: With the same period both have the same ω, so the speed depends only on the radius.

Teacher's Note:
1. The equal masses are not needed.
2. Start from ω = 2π/T and then use v = rω.

Question 2. Two bodies in circular paths of radii 1 : 3 take same time to complete their circles. Find the ratio of their linear speeds.
Answer:
Same time means the same ω, so v = rω gives v₁ : v₂ = r₁ : r₂.
Ratio = 1 : 3.
In simple words: The larger circle is covered at three times the speed.

Teacher's Note:
1. Equal time for a full circle means equal angular speed.
2. The book gives the answer 1 : 3, which agrees.

Question 3. In a circular path of radius 3m, a mass of 8kg moves with a constant speed of 20 ms^–1. Find the angular speed in radian/sec.
Answer:
ω = v ÷ r = 20 ÷ 3 = 20/3 rad/s ≈ 6.67 rad/s.
The mass is not needed.
In simple words: Divide the linear speed by the radius.

Teacher's Note:
1. Use ω = v/r.
2. The 8 kg is extra data.

Question 4. The wheel of a cycle of radius 25cm is moving with a speed 20ms^–1. Calculate the angular velocity of the wheel
Answer:
r = 25 cm = 0.25 m.
ω = v ÷ r = 20 ÷ 0.25 = 80 rad/s.
In simple words: A point on the rim at 20 m/s on a 0.25 m radius turns 80 radians each second.

Teacher's Note:
1. Change cm to m first.
2. The book's key says 28 rad/s, which is wrong (28 belongs to the 50 cm, 14 m/s question A-3).

Question 5. An air craft completes a horizontal loop of radius 2 km with a uniform speed of 600 kmh^–1. Find the angular velocity of the air craft.
Answer:
v = 600 km/h = 600 × 1000 ÷ 3600 = 500/3 m/s ≈ 166.7 m/s; r = 2 km = 2000 m.
ω = v ÷ r = (500/3) ÷ 2000 = 1/12 rad/s ≈ 0.083 rad/s.
In simple words: Convert to m/s and metres, then divide speed by radius.

Teacher's Note:
1. Alternatively ω = 600 ÷ 2 = 300 rad/h, and 300 ÷ 3600 = 1/12 rad/s.
2. Be careful with the conversion 1 km/h = 1/3.6 m/s.

Question 6. Two bodies in circular paths of radii 1 : 2 take same time to complete their circles. Find the ratio of their linear speeds.
Answer:
Same time means the same angular speed ω.
v = rω, so v₁ : v₂ = r₁ : r₂ = 1 : 2.
In simple words: Same ω, so speed is in the ratio of the radii: 1 : 2.

Teacher's Note:
1. The printed question has a typo ("Find the The ratio ... is :"); the answer in the book is 1 : 2.
2. The same question appears as MCQ A-11.

Question 7. Write down the units of angular displacement, angular velocity and angular acceleration.
Answer:
Angular displacement: radian (rad); it can also be given in degree or revolution.
Angular velocity: radian per second (rad/s).
Angular acceleration: radian per second squared (rad/s²).
In simple words: Angle is in radian, angular velocity in rad/s and angular acceleration in rad/s².

Teacher's Note:
1. Each time we divide by one more second.
2. Radian is dimensionless, so ω has the dimension of 1/time.

Question 8. The wheel of a cycle of radius 50 cm is moving with a speed 14 ms^–1. Calculate the angular velocity of the wheel.
Answer:
r = 50 cm = 0.5 m.
ω = v ÷ r = 14 ÷ 0.5 = 28 rad/s.
In simple words: Divide 14 m/s by 0.5 m to get 28 rad/s.

Teacher's Note:
1. Convert 50 cm into 0.5 m.
2. The answer matches the book key.

Question 9. An air craft completes a horizontal loop of radius 1 km with a uniform speed of 900 kmh^–1. Find the angular velocity of the air craft.
Answer:
v = 900 km/h = 900 × 1000 ÷ 3600 = 250 m/s; r = 1 km = 1000 m.
ω = v ÷ r = 250 ÷ 1000 = 0.25 rad/s.
In simple words: 250 m/s on a 1000 m radius gives 0.25 rad/s.

Teacher's Note:
1. Divide km/h by 3.6 to get m/s.
2. Keep speed and radius in the same length unit.

Question 10. An artificial satellite takes 90 minutes to complete its revolution around the earth. Calculate the angular velocity of the satellite.
Answer:
T = 90 min = 90 × 60 = 5400 s.
ω = 2π ÷ T = 2π ÷ 5400 = π/2700 rad/s ≈ 1.16 × 10^–3 rad/s.
In simple words: One revolution (2π rad) in 5400 s gives π/2700 rad/s.

Teacher's Note:
1. Use ω = 2π/T with T in seconds.
2. The answer is left in terms of π, as in the book.

Question 11. Define the time period and find the relation between v and ω.
Answer:
Time period (T) is the time taken to complete one full revolution (one round) of the circle.
In one period the body covers the circumference 2πr, so v = 2πr/T.
Also ω = 2π/T, so T = 2π/ω. Substituting in v = 2πr/T gives v = rω (or ω = v/r).
In simple words: The time for one round is the period T, and linear speed equals radius times angular speed.

Teacher's Note:
1. Mention both formulas: ω = 2π/T and v = rω.
2. Frequency f = 1/T, so ω = 2πf.