A person is riding a bike and its wheels have an angular velocity of +20.0 rad/s. then the brakes are applied and the bike is brought to a stop. during braking, the angular displacement of each wheel is +15.92 revolutions.
(a) how much time does it take the bike to come to rest?
(b) what is the angular acceleration in (rad/s^2) of each wheel?
2007-06-18 12:34:05 · 2 answers · asked by chly1459 1 in Science & Mathematics ➔ Physics
First you have to convert the wheel's angular displacement to radians by multiplying by 2π. Therefore, 15.92 x 2π = 100 rad. Use ΔΘ=1/2(ω_initial+ω_final)t to solve for t. ω_final is zero because the wheel comes to rest. Time is 2*100/20 = 10 seconds. To find the angular acceleration, you can use ΔΘ=1/2αt^2. The angular acceleration is 2*100/10^2 = -2 rad/s^2. The acceleration is negative because you are slowing down.
2007-06-18 12:51:05 · answer #1 · answered by Jeff 3 · 1⤊ 0⤋
each 2d, the shaft beneficial factors 3.00 rad/s in greater angular velocity, so the angular velocity after 18 seconds is: 3.00 rad/s^2 * 18 s = fifty 4 rad/s. to stumble on the whole perspective of rotation after 18 seconds, you combine the speed with admire to time, from t=0 to t=18 seconds: imperative((3 rad/s)*t, t, 0, 18) = (3 rad/s)*t^2 / 2, evaluated from 0 to 18 = 486 rad
2016-09-28 01:30:33 · answer #2 · answered by ? 4 · 0⤊ 0⤋