A bug is swimming laps at constant speed around outside edge of birdbath with radius 30 cm. it takes 5 s to compete one lap.

Question

A) find avg acceleration as it moves from top of the circle to bottom (12 and 6 on a clock)

 

B) the inst acceleration at (9pm) reference

Answer 1

Hi Piper!

 

So, the bug moves at a constant speed around the rim of the circle, covering one complete circumference of the circular birdbath in 5 s.  This means the bug's speed can be calculated using:

 

d = vt (the constant-velocity equation of motion)

 

d = circumference of birdbath = 2π(0.3 m) = 0.6π m

t = 5 s

 

Thus, 

 

v = d/t = 0.6π/5 = 3π/25 m/s = 0.377 m/s

 

Since this is constant-velocity motion around the circle, it is uniform circular motion.  Therefore, the only acceleration on the bug at any instant is the centripetal acceleration, point toward the center of the circle.  Thus, at the 9 pm position, the instantaneous acceleration is:

 

a_c = v²/r

 

a_c = (0.377 m/s)²/0.3 = 0.474 m/s², toward the center of the birdbath

 

The average acceleration is always defined as:

 

a_avg = (v-v_o)/Δt

 

where

 

v = final velocity (vector)

v_o = initial velocity (vector)

 

Whether the bug is circling the bath clockwise or counterclockwise, its velocity vector will have the same magnitude at the 6 and 12 o'clock positions, because its speed is constant.  But the directions will be opposite (one will be the negative of the other, since negative for a vector means opposite direction).  The magnitude is 0.377 m/s, as computed above.

 

Also, if the bug moves at constant speed, and takes 5 s to circle the birdbath once, it takes 2.5 s to go from the 6 o'clock to the 12 o'clock positions.  

 

So the average acceleration magnitude will be:

 

a_avg = (0.377 - (-0.377))/(2.5) = 0.754/2.5 = 0.302 m/s², along the line tangent to the circle at 6 or 12 o'clock (which way along that line depends on whether the bug is moving clockwise or counterclokwise)

 

Note that this is different from the instantaneous acceleration at 9 o'clock.  This is because it is effectively the "average" of all the centripetal acceleration vectors at every little instant between the 6 and 12 o'clock positions.  This vectors do not all point the same way, so they partly cancel out, making the average velocity somewhat less than the instantaneous velocity at any one point.

 

I hope this helps!  Please let me know if you have any other questions about this or similar problems.