Question

For no apparent reason, a poodle is running counter-clockwise at a constant speed of 4.90 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velocity vector at time t1, and let v⃗ 2 be the velocity vector at time t2. Consider Δv⃗ =v⃗ 2−v⃗ 1 and Δt=t2−t1. Recall that a⃗ av=Δv⃗ /Δt. Hint: It may be helpful to assume that at time t1, the poodle is on the x-axis, i.e., that the velocity vector v¯1 points along the y-axis.

A)For Δt = 0.5 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.

b)For

Δt = 4×10⁻² s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.

Answer 1

Suppose initially v₁ was along y axis ( to that to calculate the direction of a_avg relative to v₁ we need to calculate only the angle made by the relative acceleration with X and Y axis

Now initially velocity was along positive y axis

So v₁ = 4.9 j m/s

After time t the position vector of particle will be rotated by angle A and velocity vector will also be rotated by same angle A along counterclockwise sense . ( ie now the velocity vector will make angle A with positive Y axis and angle (90-A ) with negative X axis

We can write

V₂ = -4.9 sin(A) i + 4.9 cos(A) j m/s

And angle A can be calculated as

A = V*T /r = 4.9t / 2.9 radian ( V speed , T time and r radius )...............Eq.1

(A)

Avg. Acceleration = v2- v1 /∆t = -4.9 sin(A) i - 4.9{1- cos(A)} j / 0.5 meter per second square = -9.8sin(A)i - 9.8 (1-cosA)j

But the direction of relative to v1 means we have to calculate angle made by avg.acceleration with v1 (or with Y axis because v1 is || to Y axis)

And A = 4.9/5.8 radian = 0.84 radian

Putting all the values we get angle the angle made by avg. Acceleration with positive Y axis will be

B= $\Pi -\tan^{-1}\frac{sinA}{1-cosA} where A = 0.84radian$

(B) the method will be same for solving this part but the difference is only that now since T has been changed so the value of angle A will be different

Here A= 4.9*0.04/2.9 = 0.0676 radian (very small value so we can assume that. sinA~ A and 1- cosA ~ A/2)

There for putting those values in the euation angle made by acceleration with positive Y axis is

$\Pi -\tan^{-1}\frac{sinA}{1-cosA} = \Pi - \tan^{-1}2$