The speed of a point on a rotating turntable,
which is 0.172 m from the center, changes at a
constant rate from rest to 0.531 m/s in 2.19 s.
At t1 = 1.9 s, find the magnitude of the
tangential acceleration.
Answer in units of m/s2.
At t1 = 1.9 s, find the magnitude of the total
acceleration of the point.
Answer in units of m/s2.
I just did
oh okay sorry I'm not the best at this and iv been super stuck >< thank you for the help
Oh boy, we're really spinning into action with this question! To find the magnitude of the tangential acceleration, we can use the formula for acceleration on a rotating object:
a = (v - u) / t
Where:
a = tangential acceleration
v = final velocity
u = initial velocity
t = time taken
Given that the initial velocity, u, is 0 m/s (since the point starts from rest), the final velocity, v, is 0.531 m/s, and the time taken, t, is 2.19 s, we can plug in these values:
a = (0.531 m/s - 0 m/s) / 2.19 s
Simplifying this expression gives us:
a = 0.531 m/s / 2.19 s
Now let's calculate:
a ≈ 0.242 m/s²
Therefore, the magnitude of the tangential acceleration at t1 = 1.9 s is approximately 0.242 m/s². Keep spinning those numbers!
To find the magnitude of the tangential acceleration at t1 = 1.9 s, we need to determine the change in speed (∆v) and the time interval (∆t) during that time.
Given:
Radius of the turntable (r) = 0.172 m
Initial speed (v1) = 0 m/s
Final speed (v2) = 0.531 m/s
Time interval (∆t) = t2 - t1 = 2.19 s - 1.9 s = 0.29 s
To find ∆v, we subtract the initial speed from the final speed:
∆v = v2 - v1 = 0.531 m/s - 0 m/s = 0.531 m/s
Now, we can calculate the magnitude of the tangential acceleration (at) using the formula:
at = ∆v / ∆t
Substituting the values we have:
at = 0.531 m/s / 0.29 s
Calculating the result:
at ≈ 1.83 m/s²
Therefore, the magnitude of the tangential acceleration at t1 = 1.9 s is approximately 1.83 m/s².
The angular acceleration rate is
alpha = (Angular velocity at 2.19 s)/2.19s
= (0.531/0.172)/2.19 = 1.41 rad/s^2
The tangential acceleration rate remains constant at R*alpha = 0.242 m/s^2 during the turntable acceleration. The 1.9 s time is not needed for the calculation, since alpha remains constant. Centripetal acceleration increases with the square of time.