
#1
Nov2607, 02:38 PM

P: 5

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 merrygoround 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? 



#2
Nov2607, 06:30 PM

Sci Advisor
HW Helper
P: 6,574

AM 



#3
Nov2607, 07:01 PM

P: 5

1.
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! 



#4
Nov2607, 07:28 PM

Sci Advisor
HW Helper
P: 6,574

check me pleaseAM 


Register to reply 
Related Discussions  
Please check if I did it right?  Introductory Physics Homework  1  
Anyone can Check for me?  Introductory Physics Homework  2  
Can someone check if i'm doing this right?  Introductory Physics Homework  2  
Can someone check my lab?  Introductory Physics Homework  5  
check  Introductory Physics Homework  6 