OSC An ultracentrifuge accelerates from rest to 100,000 rpm in 2 min . What is the tangential acceleration of a point 9.50cm from the axis of rotation? Your answer should be given in m/sec^2 .
To find the tangential acceleration, we need to first find the angular acceleration (α) of the ultracentrifuge.
We know the initial angular velocity (ω_0) is 0 since it starts from rest. The final angular velocity (ω) is given as 100,000 rpm, which needs to be converted to radians per second (rad/s).
1 revolution = 2π radians
100,000 rpm = 100,000 revolutions per minute
= (100,000 rev/min) * (2π rad/rev) * (1 min/60 s)
= (100,000 * 2π * 1) / 60 rad/s
≈ 10,472 rad/s
The time taken (t) is given as 2 min, which needs to be converted to seconds.
1 min = 60 s
2 min = 2 * 60 s = 120 s
Now, we can calculate the angular acceleration (α) using the formula:
ω = ω_0 + α * t
10,472 rad/s = 0 + α * 120 s
α = (10,472 rad/s) / (120 s) ≈ 87.27 rad/s^2
To find the tangential acceleration (a_t) at a point 9.50 cm from the axis of rotation, we can use the formula:
a_t = r * α
where r is the radius from the axis of rotation, given as 9.50 cm.
Converting cm to meters:
9.50 cm = 9.50/100 m = 0.095 m
a_t = 0.095 m * 87.27 rad/s^2 ≈ 8.28 m/s^2
Therefore, the tangential acceleration of a point 9.50 cm from the axis of rotation is approximately 8.28 m/s^2.