Question

a car accelerates from 0 to 72 km h in 5 seconds. If it has wheels of diameter 50cm. the angular acceleration of its wheels is?​

Answer 1

Answer:

Correct option is A)

Correct option is A)Initial speed, v=72 km/h

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/s

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 m

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π rad

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/s

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/s

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/sFrom kinematics:

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/sFrom kinematics:ω′2−ω2=2αθ (where α is angular acceleration)

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/sFrom kinematics:ω′2−ω2=2αθ (where α is angular acceleration)⟹0−6400=2α×40π

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/sFrom kinematics:ω′2−ω2=2αθ (where α is angular acceleration)⟹0−6400=2α×40π⟹α=80π−6400

Correct option is A)Initial speed, v=72 km/h⟹v=360072×103⟹v=20 m/sDistance, 2r=50 cm⟹r=25×10−2 mAngular displacement, θ=20×2π⟹θ=40π radInitial angular velocity, ω=rv⟹ω=25×10−220⟹ω=54×102⟹ω=80 rad/sFinal angular velocity, ω′=0 rad/sFrom kinematics:ω′2−ω2=2αθ (where α is angular acceleration)⟹0−6400=2α×40π⟹α=80π−6400⟹α=−25.5 rad/s2

Answer 2

Concept:

• Rotational motion
• Acceleration
• Angular acceleration

Given:

• the initial velocity of car u = 0
• the final velocity of the car v = 72 km/h = 72* 1000/3600 = 72* 5/18 = 20 m/s
• The time taken for the car to accelerate t = 5 s
• Diameter of wheels = 50 cm
• Radius of wheels = diamter/2 = 50/2 = 25 cm = 0.25 m

Find:

• The angular acceleration of the wheels

Solution:

We know the following kinematics equation:

v = u + at

v = 20 m/s, u = 0 m/s, t = 5 s

20 = 0 + a (5)

5a = 20

a = 20/5

a = 4 m/s²

We know that linear acceleration is the product of the radius and the angular acceleration

a = Rα

α = a/R

α = 4/ 0.25

α = 16 rad/ s²

The angular acceleration of the wheels is 16 rad/ s².

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