a 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 2.0 rad/s^2 over a 3.54-s interval
d = 0.5*a*t^2 = 0.5*2*3.54^2 = 12.53 Rad
To find the number of radians the tire rotates, we can use the kinematic equation for rotational motion:
θ = θ0 + ω0t + (1/2)αt^2
Where:
θ = final angle (in radians)
θ0 = initial angle (in radians)
ω0 = initial angular velocity (in rad/s)
α = angular acceleration (in rad/s^2)
t = time (in seconds)
Given:
r = 0.30 m (radius of the tire)
α = 2.0 rad/s^2 (angular acceleration)
t = 3.54 s (time interval)
First, we need to find the initial angular velocity (ω0). Since the tire starts from rest, ω0 = 0 rad/s.
Next, we can calculate the final angle (θ):
θ = ω0t + (1/2)αt^2
θ = 0 + (1/2)(2.0)(3.54)^2
θ = (1/2)(2.0)(12.5316)
θ ≈ 12.532 radians
Therefore, the 0.30-m radius automobile tire rotates approximately 12.532 radians after starting from rest and accelerating at a constant 2.0 rad/s^2 over a 3.54-s interval.
To find the number of radians the automobile tire rotates, we can use the equation:
θ = ω₀t + 0.5αt²
Where:
θ is the angular displacement (in radians)
ω₀ is the initial angular velocity (in radians/s)
α is the angular acceleration (in radians/s²)
t is the time interval (in seconds)
First, let's calculate the initial angular velocity using the formula:
ω₀ = 0 (since the tire starts from rest)
Next, substitute the values into the equation:
θ = (0) × (3.54) + 0.5 × (2.0) × (3.54)²
Simplifying:
θ = 0 + 0.5 × 2.0 × 12.5316
Calculating:
θ = 12.5316
Therefore, the automobile tire rotates approximately 12.5316 radians during the 3.54-second interval.