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Learn/ 11th Physics/ Unit 3 /Numericals

Unit 3: Circular and Rotational Motion — Numericals

11th Class Physics · Unit 3: Circular and Rotational Motion

3.1.A laser beam is directed from the Earth to the moon. The beam spreads over a diameter of 2.50 m at the moon surface. What is divergence angle of the beam? The distance of moon from the Earth is 3.8 x 10^8 m.
Given
Diameter of beam at moon's surface S = 2.50 m
Distance from Earth to Moon r = 3.8 × 108 m
Formula
Arc length relation S = rθ
Divergence angle θ = Sr
Working
Substitution θ = frac{2.50}{3.8 × 108 m
Result
Final answer θ ≈ 6.58 × 10-9 rad
3.2.A car is moving with a speed of 108 km h^-1. If its wheel has a diameter of 60 cm, find its angular speed in rad s^-1 and rev.s^-1.
Given
Speed of car v = 108 km h-1 = 108 × 10003600 ms-1 = 30 ms-1
Diameter of wheel D = 60 cm = 0.6 m
Radius of wheel r = D2 = 0.3 m
Formula
Angular speed omega = vr
Working
Angular speed calculation omega = 300.3 = 100 rad.s-1
Convert rad/s to rev/s omega = 1002pi rev.s-1 = 1002 × 3.14 rev.s-1 ≈ 15.92 rev.s-1
Result
Final answers omega ≈ 100 rad.s-1 or approximately 15.92 rev.s-1
3.3.An electric motor is running at 1800 rev min^-1. On switching off, it comes to rest in 20 s. If angular retardation is uniform, find the number of revolutions it makes before stopping.
Given
Initial angular speed omegai = 1800 rev.min-1 = 180060 rev.s-1 = 30 rev.s-1
Time to stop t = 20 s
Final angular speed omegaf = 0 rev.s-1 (comes to rest)
Formula
Average angular velocity omegaav = omegai + omegaf2 = 30 + 02 = 15 rev.s-1
Total revolution using average velocity θ = omegaav × t = 15 × 20 = 300 revolution
Result
Final answer θ = 300 revolutions
3.4.A string 0.5 m long holding a stone can withstand maximum tension of 35.6 N. Find the maximum speed at which a stone of 0.5 kg mass can be whirled with it in a vertical circle.
Given
Length of string r = 0.5 m
Maximum tension T = 35.6 N
Mass of stone m = 0.5 kg
Tension at bottom of circle T = mg + mv2r (at the bottom of the circle), the tension will be maximum
Working
Substituting values 35.6 = 0.5 × 9.8 + 0.5 × v20.5
Simplification 36.6 = 4.9 + v2
Solving for v^2 35.6 - 4.9 = v2
Result v2 = 30.7
Result
Maximum speed v = 5.54 ms-1
3.5.The flywheel of an engine is rotating at 2100 rev min^-1 when the power source is shut off. What torque is required to stop it in 3 minutes? The moment of inertia of the flywheel is 36 kg m^2.
Given
Initial speed omegai = 2100 rev.min-1
Time to stop t = 3 min = 180 s
Moment of inertia I = 36 kg m2
Final speed omegaf = 0
Convert to rev/s omegai = 210060 = 35 rev.s-1
Convert to rad/s omegai = 35 × 2pi = 70pi rad.s-1
Angular acceleration α = omegaf - omegait = 0 - 70pi180 rad.s-1
Angular acceleration simplified α = -70pi180 rad.s-1
Formula
Torque formula tau = Iα
Working
Torque calculation tau = 36 × left(-70pi180right)
Result
Final answer tau ≈ -43.98 Nm ≈ -44 Nm
3.6.What is the moment of inertia of a 200 kg sphere whose diameter is 60 cm?
3.7.A satellite is orbiting the Earth at an altitude of 200 km. Assuming the Earth's radius is 6400 km, calculate the orbital speed of the satellite.
Given
Altitude of satellite h = 200 km = 2 × 105 m
Radius of Earth R = 6400 km = 6.4 × 106 m
Orbital radius r = R + h = 6.4 × 106 + 2 × 105 = 6.6 × 106 m
Formula
Orbital speed v = sqrt{GMr
Working
Substitution with values v = sqrt{frac{6.67 × 10-11 × 6 × 1024{6.6 × 106}
Calculation v = sqrt{frac{6.67 × 10-11 × 6 × 1024{6.6 × 106} = 7.78 × 103 ms-1
Result
Final answer v = 7.78 kms-1
3.8.A space station has a radius of 20 m and rotates at an angular velocity of 0.5 rads^-1. What is the artificial gravity experienced by the astronauts on the space station?
Given
Radius of space station r = 20 m
Angular velocity omega = 0.5 rad.s-1
Artificial gravity formula g = romega2
Working
Substitution g = 20 × (0.5)2
Calculation g = 20 × 0.25 = 5 ms-2
Result
Final answer g = 5 ms-2
3.9.A bicycle wheel has an angular momentum of 10 kg m^2s^-1 and angular velocity of 2 rads^-1. Find the value of its moment of inertia.
Given
Angular momentum L = 10 kg m2s-1
Angular velocity omega = 2 rad.s-1
Moment of inertia = ? I = ?
Formula
Relation between L, I, and ω L = Iomega
Solving for I I = Lomega = 102
Result
Final answer I = 5 kg m2
3.10.A diver comes off a board with arms straight up and legs straight down, giving him a moment of inertia of 18 kg m^2 about his rotation axis. Then tucks into a small ball, decreasing his moment of inertia to 3.6 kg m^2. While tucked, he makes two complete rotations in 1.0 second. If he had not tucked at all, how many revolutions would he have made in 1.5 s from board to water?
Given
Initial moment of inertia I1 = 18 kg m2
Tucked moment of inertia I2 = 3.6 kg m2
Time from board to water t = 1.5 s
Tucked angular velocity (2 revolutions in 1.0 s) omega2 = 2 rev/s
Formula
Conservation of angular momentum I1omega1 = I2omega2
Solving for initial ω omega1 = I2I1omega2 = 3.618 × 2 rev.s-1
Initial angular velocity omega1 = 0.2 × 2 = 0.4 rev.s-1
Number of revolutions Revolutions (θ) = omega × t
Working
Calculation θ = 0.4 × 1.5 = 0.6 rev
Result
Final answer He would have made 0.6 rev