series of questions that I need to know if my logic behind the answers is correct. A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge and the girl is closer to the center. Who has the 1. greater angular displacement a) boy b) girl c) both have the same angular displacement I figured it would be the girl because angular displacement is inversely proportional to the radius because of arc length=(radius)(angular disp). 2. greater angular speed a) boy b) girl c) both have the same angular velocity I thought it would be the girl again since v=rw so if the radius decreases, the angular velocity increases. 3. greater linear speed a) boy b) girl c) both have the same linear speed I'm unsure of this one. I'd probably choose the boy since I think he has a greater displacement, so the velocity would be higher. 4. greater centripetal acceleration a) boy b) girl c) both have the same I thought that the answer would be the girl because a(sub c)=(v^2)/r so if the radius went up like the boy's then the acceleration would go down. I'm not sure if I used the right equation for this one... 5. greater tangential acceleration a) boy b) girl c) both have the same I completely guessed on this one and said they had the same since tangential acceleration = r(ang accel). Yeah, I really don't get this one. Are these right?
Angular displacement is the total angle through which the body has rotated. Does that angle depend on radius? Angular speed is [itex]\omega = v/r[/itex] and [itex]v = 2\pi r/T[/itex]. So [itex]\omega = v/r = 2\pi/T[/itex]. T is the period of rotation. Does that depend on radius? You actually got this right but I don't think you understand why. It follows from [itex]v = \omega r[/itex]. Since [itex]\omega[/itex] is the same for each, how does the tangential speed change as r increases. Use [itex]v = \omega r[/itex] so [itex]a = v^2/r = \omega^2 r[/itex]. Since [itex]\omega[/itex] is the same for both, how does a change with r? Tangential acceleration is [itex]a = \alpha r = \dot{\omega} r[/itex]. If [itex]\omega[/itex] is constant, what is [itex]\dot{\omega}[/itex]? So what is [itex]\alpha[/itex]? a? AM
1. The angle of a circle wouldn't depend on the radius, so that would mean that they both have the same angular displacement. 2. If w = 2(pi) / T then it doesn't depend on the radius either, so the answer would be they're both the same again, right? 3. Huzzah! The one I got right! Your explanation does make a lot more sense though. 4. Since w is the same for both, that would mean that as the radius increases, so does the centripetal acceleration, making the boy have the greater v. 5. I'm not sure that I really understand your explanation of the last one. Would the equation you used cause the answer to be the boy because the radius and the tangential acceleration are directly proportional? Your reply was amazing. It helped me so much, thanks!
No. It is simple: Angular acceleration is the rate of change of angular speed. If the angular speed is constant (for both), what is the angular acceleration (for both)? AM