Compute the speed of the tip of the second hand

[smoki]

I think this is a simple question but I'm not sure if I'm doing it right, this is what I have so far...


A watch has a second hand 2.0 cm long.


a) Compute the speed of the tip of the second hand.

C = 2pi(r) r = 2.0cm = 0.02m - leave as cm or convert to m?

V = [2pi(0.02m)]/60s = 2.1 x 10^-3 m/s - divide by 60s?

b) What is the velocity of the tip of the second hand at 0.0s ? At 15 seconds?

well 0s=60s right...? if so then the answer is the same as a)

for 15s V=[2pi(0.02m)]/15s = 8.3 x 10^-3 m/s

c) Compute its change in velocity between 0.0 and 15 seconds.

?

d) Compute its average vector acceleration between 0.0 and 15 seconds.

I think I could answer this if I could figure out c)

 

[FredGarvin]

Part A looks correct.

For part B, remeber that velocity is a vector...amplitude and direction. Therefore the velocity at 0/60 sec is (using the i designator as a unit vector in the +x direction) $$\vec{v} = 2.1 x 10^{-3} \hat{i}$$. At 15 seconds, the direction is going to be down (using j as the unit vector in the +y direction), in the -j direction with the same magnitude.

Does that point out what you need to do for the rest of the problems?

 

[thepatientmental]

a) To compute the speed of the tip of the second hand, we can use the formula v = 2πr/t, where r is the radius (length) of the second hand and t is the time in seconds. In this case, r = 2.0 cm = 0.02 m and t = 60 seconds (as the second hand completes one full rotation in 60 seconds). So the speed of the tip of the second hand is: v = (2π*0.02m)/60s = 0.0021 m/s.

b) The velocity of the tip of the second hand at 0 seconds is the same as the answer in part a), which is 0.0021 m/s. At 15 seconds, the second hand has completed 1/4th of a rotation, so we can use the formula v = (2π*0.02m)/(1/4*60s) = 0.0083 m/s.

c) The change in velocity between 0 and 15 seconds can be calculated by taking the difference between the velocities at these two times. So the change in velocity is: 0.0083 m/s - 0.0021 m/s = 0.0062 m/s.

d) To compute the average vector acceleration, we need to know the initial and final velocities as well as the time interval. In this case, the initial velocity at 0 seconds is 0.0021 m/s and the final velocity at 15 seconds is 0.0083 m/s. The time interval is 15 seconds. So the average vector acceleration is: (0.0083 m/s - 0.0021 m/s)/15s = 0.00028 m/s^2.