A ceiling fan with 80-cm-diameter blades is turning at 60 rpm....

Question

A ceiling fan with 80-cm-diameter blades is turning at 60 rpm. Suppose the fan coasts to a stop 25 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off?

Accepted Answer

Hi everyone. In this practice problem, we are being asked to calculate the linear velocity of a point on the disc's outer edge 15 seconds after the fuse blows. In this case, we will have a disc of a standard machine rotating at a constant speed of R PM. And the discs diameter is going to be 10 centimeter. All of a sudden the machine fuse blows and the disc takes seconds to stop. We're being asked to calculate the linear velocity of a point on the disc outer edge 15 seconds after the fuse blows. And the options given are a 0.1 m per second. B 2.1 m per second, C 8.5 m per second and D 33.5 m per second. So in order for us to um solve this problem, we will assume that the disk will accelerate at a constant rate. So the rate of change of angular velocity of a particle rotating at an angle in a circular orbit is going to be called the angular acceleration. So what we are interested at in looking is going to be the angular acceleration first in order for us to be able to use the angular acceleration in a kinematic equation to get the angular velocity. And from there, we'll get the linear velocity. So the kinematic equation that we will finally use is going to be omega F equals omega I plus alpha T. What we are interested at in looking is the omega F which is the angular velocity after the 15 seconds after the fuse blows. And then from there, we know that um Omega will be directly correlated to the linear velocity V by this following equation. So V will equals two R multiplied by omega or in this case, Omega F and V is going to be our V F which is the velocity after 15 seconds after diffuse blows. In order for us to use our kinematic equation, what we need to find first is the angular acceleration. So the angular acceleration will be able to be calculated through alpha equals Omega two minus omega one divided by delta T. Uh we are given to, we are given the Omega two which is going to be zero radiance per second in this case. And the Delta T is going to be the 36 seconds that it takes for it to actually comes to a full stop. The omega one given is going to be 700 R PM, which we want to convert into radiant per seconds just so that it's easier for us to actually use in the following calculation 700 R PM is equals to 700 revolution per minute. We want to multiply that by two pi radiant per one revolution. And we want to multiply the uh that as well with one minute for every 60 seconds just like. So and that will actually give us omega one value of 73.3 radiant per seconds just like. So, so in order for us to calculate the angular acceleration, we wanna look at the case where the actual machine is rotating from the 700 R PM all the way to the stop because that is what is given in the problem statement or the given information in the problem statement. So the angular acceleration within those 36 seconds will be the same with the angular acceleration up until the 15 seconds because we are assuming that the deceleration is going at a constant rate throughout. So now that we have all the necessary information to find the angular acceleration, we will input everything into our equation. So Omega two is zero radiance per second. Omega one is 73.3 radiant per seconds. Delta T is going to be 36 seconds and it will give us an angular acceleration of -2.04 radiance per seconds just like that. Now, we can actually plug everything into our um kinematic equation which from the or in the kinematic equation, Omega I is going to equals to Omega one which is going to be 73.3 radiance per second. So Omega F equals Omega I plus alpha, T with Omega, I being 73.3 radium per seconds plus alpha is going to be negative two point oh four radium per seconds. And the time is going to be the 15 seconds because that is what we are interested at looking or what's being asked in the problem statement, which is the linear velocity 15 seconds after the fuse blows. So this will actually give us an omega f value of 42.7 radiant per second. And now we can use that um angular velocity to calculate the linear velocity at a point on the disc's outer edge 15 seconds after the fuse blows. So we want to utilize P F equals R multiplied by W or Omega F and the R the diameter given for the disc is going to be 10 centimeter. So that means the R is going to be five centimeters And that will essentially be 0.05 m. So we want to utilize that into our recreation. D R is going to be 0.5 m and the Omega F is going to be 42.7 radiance per seconds. Dell will give us the linear velocity of 2.1 m per seconds. So the linear velocity off a point on the disc's outer edge 15 seconds after the fuse blows is going to be V F which is 2.1 m per second. And that will correspond to option B in our problem statement. So option B is going to be the answer to this particular practice problem With a linear velocity of 2.1 m/s. So if you guys still have any sort of confusion on this particular practice problem, please make sure to check out our other lesson videos and that'll be all for this one. Thank you.

A ceiling fan with 80-cm-diameter blades is turning at 60 rpm. Suppose the fan coasts to a stop 25 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off?

Key Concepts

Angular Velocity

Linear Speed

Deceleration

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