
#1
Jun1009, 08:00 PM

P: 1

The minute hand of a clock is 5.5 cm long. What is average velocity vector for the tip of the hand during the interval from the hour to 20 minutes past the hour, expressed in a coordinate system with theyaxis toward noon and xaxis toward 3 o'clock? (Answer in terms of ihat and jhat components please)




#2
Jun1009, 10:01 PM

P: 51

Everyone here has helped me so much, so I will do the same. Ok you know that in 20 minutes, the minute hand moves 1/3 of the circle. Therefore it moves 120 degrees. Drawing a picture helps here. Now you have a isosceles triangle with sides 5.5 and vertex angle 120. Solve to get the other side and divide by 20 for an answer in cm/min.




#3
Jun1109, 12:25 AM

P: 69

Velocity is equal: v=r x ω
ω is the angular velocity of the minute hand. You know that the minute hand makes 2pi in one minute (60 seconds). So angular velocity is equal: ω= 2*pi/60 = 1/60 pi. Velocity (perpendicular on the minute hand) is equal to :v=r x ω=0,055 * 1/60 pi. Velocity in i & j is v= v_{x}i + v_{y}j. v_{x}=v*cos60 v_{y}=v*sin60 (I have attached a photo) 


Register to reply 
Related Discussions  
Constant velocity questions (average speed vs. average velocity)  Introductory Physics Homework  2  
Average velocity and average speed problem.  Introductory Physics Homework  12  
Vector components on the tip of hand on a clock  Introductory Physics Homework  3  
Average Speed/Average Velocity Problem  Introductory Physics Homework  4 