At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 3.0 rad/s2 until a circuit breaker trips at t = 2.00 s. From then on it turns through 42 rad as it coasts to a stop at constant angular acceleration
Incomplete.
To find the time it takes for the grinding wheel to come to a stop, we can use the formula for angular velocity:
ω = ω_0 + αt,
where ω is the final angular velocity, ω_0 is the initial angular velocity, α is the angular acceleration, and t is the time.
Given:
ω_0 = 24.0 rad/s (initial angular velocity)
α = 3.0 rad/s² (constant angular acceleration)
t = 2.00 s (time at which the circuit breaker trips)
θ = 42 rad (angular displacement)
First, we need to find the final angular velocity when the circuit breaker trips. We can use the formula:
ω = ω_0 + αt.
Substituting the given values, we have:
ω = 24.0 rad/s + 3.0 rad/s² * 2.00 s
= 24.0 rad/s + 6.0 rad/s
= 30.0 rad/s.
Next, let's find the time it takes for the grinding wheel to come to a stop. We can use the formula:
θ = ω_0t + (1/2)αt².
Since we want to find the time it takes for the wheel to stop, we set ω = 0 and solve for t:
θ = ω_0t + (1/2)αt²
42 rad = 30.0 rad/s * t + (1/2) * 3.0 rad/s² * t²
42 rad = 30.0t rad/s + 1.5t² rad/s²
1.5t² + 30.0t - 42 = 0.
This is a quadratic equation that we can solve for t using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation.
In our case, a = 1.5, b = 30.0, and c = -42. Substituting these values into the quadratic formula, we have:
t = (-(30.0) ± √((30.0)² - 4 * 1.5 * (-42))) / (2 * 1.5).
Simplifying this expression, we get two possible solutions for t:
t₁ = (-30.0 + √(900 + 252)) / 3
t₁ = (-30.0 + √1152) / 3
t₁ ≈ 3.014 s.
t₂ = (-30.0 - √(900 + 252)) / 3
t₂ = (-30.0 - √1152) / 3
t₂ ≈ -23.014 s.
Since time cannot be negative in this context, we discard the negative solution. Therefore, the grinding wheel takes approximately 3.014 seconds to come to a stop after the circuit breaker trips.