At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration of –0.40 rad/s^2 has an angular velocity of 1.5 rad/s and an angular position of 2.3 rad. What is the angular position of the wheel
at t = 2.0 s? Ans: Theta final=4.5 rad
How do you solve step by step. Thanks
èf = èi+ùi*t+1/2*á*t^2
èf = 2.3rad+(1.5 rad⁄s(2.0s)+0.5(-0.40 rad⁄s^2)(2.0s)
èf = 2.3 rad + 3.0 rad - 0.8 rad
èf = 4.5 rad
Thanks! Bob Pursley, This helped me solve it and here is the step by step break down.
angular position= 1/2 acceleraiton*time^2+ wi*time + initial position
Well, let's go step by step then!
First, we need to find the angular position of the wheel at t = 2.0 s. To do that, we need to use the kinematic equation for rotational motion:
θ(t) = θ(initial) + ω(initial) * t + (1/2) * α * t^2
where θ(t) is the angular position at time t, θ(initial) is the initial angular position, ω(initial) is the initial angular velocity, α is the angular acceleration, and t is the time.
Given that θ(initial) = 2.3 rad, ω(initial) = 1.5 rad/s, α = -0.40 rad/s^2, and t = 2.0 s, we can plug in these values into the equation:
θ(2.0 s) = 2.3 rad + 1.5 rad/s * 2.0 s + (1/2) * (-0.40 rad/s^2) * (2.0 s)^2
Simplifying the equation gives us:
θ(2.0 s) = 2.3 rad + 3.0 rad - 0.40 rad/s^2 * 2.0 s^2
θ(2.0 s) = 2.3 rad + 3.0 rad - 0.80 rad
θ(2.0 s) = 4.5 rad
So, the angular position of the wheel at t = 2.0 s is 4.5 rad.
Remember, a clown's advice is to always have fun and not take everything too seriously, including solving physics problems!
To find the angular position of the wheel at t = 2.0 s, we can use the following equation:
θf = θi + ωi * t + (1/2) * α * t^2
Given:
θi = 2.3 rad (initial angular position)
ωi = 1.5 rad/s (initial angular velocity)
α = -0.40 rad/s^2 (angular acceleration)
t = 2.0 s (time)
Step 1: Calculate the first term of the equation (θi)
θi = 2.3 rad
Step 2: Calculate the second term of the equation (ωi * t)
ωi * t = 1.5 rad/s * 2.0 s = 3.0 rad
Step 3: Calculate the third term of the equation ((1/2) * α * t^2)
(1/2) * α * t^2 = (1/2) * (-0.40 rad/s^2) * (2.0 s)^2 = -0.80 rad
Step 4: Add up the three terms to find the final angular position (θf)
θf = θi + ωi * t + (1/2) * α * t^2
= 2.3 rad + 3.0 rad - 0.80 rad
= 4.5 rad
Therefore, the angular position of the wheel at t = 2.0 s is 4.5 rad.
To solve this problem, you can use the equations of rotational motion.
The equation that relates angular position, final angular velocity, initial angular velocity, angular acceleration, and time is:
θf = θi + ωit + 1/2 αt^2
Where:
θf is the final angular position
θi is the initial angular position
ωi is the initial angular velocity
α is the angular acceleration
t is the time
Given:
θi = 2.3 rad (initial angular position)
ωi = 1.5 rad/s (initial angular velocity)
α = -0.40 rad/s^2 (angular acceleration)
t = 2.0 s (time)
Now, plug in the given values into the equation:
θf = 2.3 rad + (1.5 rad/s)(2.0 s) + 1/2 (-0.40 rad/s^2)(2.0 s)^2
Simplifying the equation:
θf = 2.3 rad + 3.0 rad + (-0.4 rad/s^2)(2.0 s)^2
θf = 2.3 rad + 3.0 rad + (-0.4 rad/s^2)(4.0 s^2)
θf = 2.3 rad + 3.0 rad + (-0.4)(4.0) rad
θf = 2.3 rad + 3.0 rad - 1.6 rad
θf = 4.5 rad
Therefore, the angular position of the wheel at t = 2.0 s is 4.5 rad.