A motorcyclist is traveling along a road and accelerates for 4.45 s to pass another cyclist. The angular acceleration of each wheel is +6.35 rad/s2, and, just after passing, the angular velocity of each wheel is +75.6 rad/s, where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?
v= 75.6
v=u+at
at= 6.35 x 4.45= 28.26
75.6 - 28.26 = u = 47.34
v^2=u^2+2as
75.6^2 = 47.34^2+(2 x 6.35 x s)
5715.36-2241.08=12.7 x s
3474.27/12.7= s (answer is in rad)
Why did the motorcycle take up comedy? Because it wanted to become a wheel-y funny vehicle.
Now, let's calculate the angular displacement of each wheel during this time. The formula for angular displacement is:
θ = ωi * t + (1/2) * α * t^2
Where:
θ is the angular displacement
ωi is the initial angular velocity
α is the angular acceleration
t is the time
Given:
ωi = 0 rad/s (initially at rest)
α = +6.35 rad/s^2
t = 4.45 s
Let's calculate the angular displacement for each wheel:
For the first wheel:
θ1 = ωi * t + (1/2) * α * t^2
θ1 = 0 * 4.45 + (1/2) * 6.35 * (4.45)^2
θ1 = 0 + (1/2) * 6.35 * 19.8025
θ1 = (1/2) * 125.518375
θ1 = 62.7591875 rad
For the second wheel:
θ2 = ωi * t + (1/2) * α * t^2
θ2 = 0 * 4.45 + (1/2) * 6.35 * (4.45)^2
θ2 = 0 + (1/2) * 6.35 * 19.8025
θ2 = (1/2) * 125.518375
θ2 = 62.7591875 rad
So, the angular displacement of each wheel during this time is approximately 62.76 radians.
To calculate the angular displacement of each wheel, we can use the following formula:
θ = ω_i * t + (1/2) * α * t^2
Where:
θ is the angular displacement,
ω_i is the initial angular velocity,
t is the time, and
α is the angular acceleration.
Given:
ω_i = 0 rad/s (as it is not mentioned)
t = 4.45 s
α = +6.35 rad/s^2
ω_f = +75.6 rad/s
First, let's calculate the initial angular velocity:
ω_f = ω_i + α * t
ω_i = ω_f - α * t
ω_i = 75.6 rad/s - 6.35 rad/s^2 * 4.45 s
ω_i = 75.6 rad/s - 28.2575 rad/s
ω_i = 47.3425 rad/s
Now, let's calculate the angular displacement of each wheel:
θ = ω_i * t + (1/2) * α * t^2
θ = 47.3425 rad/s * 4.45 s + (1/2) * 6.35 rad/s^2 * (4.45 s)^2
θ = (210.714875 rad) + (0.5 * 253.046625 rad)
θ = 210.714875 rad + 126.5233125 rad
θ = 337.2381875 rad
Therefore, the angular displacement of each wheel during this time is 337.2381875 rad.
To find the angular displacement of each wheel, we need to use the formula:
θ = ω_i * t + (1/2) * α * t^2
Where:
- θ is the angular displacement
- ω_i is the initial angular velocity
- α is the angular acceleration
- t is the time
Given:
- α = +6.35 rad/s^2 (counterclockwise direction)
- ω_i = 0 rad/s (initial angular velocity)
- t = 4.45 s
For each wheel, the initial angular velocity is zero (as stated in the problem), so we only need to calculate the second term of the equation.
For the first wheel:
θ = 0 * 4.45 + (1/2) * 6.35 * (4.45)^2
θ = 0 + (1/2) * 6.35 * 19.8025
θ = 0 + 63.2780375
θ = 63.278 rad
For the second wheel:
θ = 0 * 4.45 + (1/2) * 6.35 * (4.45)^2
θ = 0 + (1/2) * 6.35 * 19.8025
θ = 0 + 63.2780375
θ = 63.278 rad
Therefore, the angular displacement of each wheel during this time is 63.278 radians (counterclockwise direction).