Question

Deriving centripetal acceleration (shortened) 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. Part A For Δt = 0.5 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av. Express your answer using four significant figures. aav = m/s2 SubmitMy AnswersGive Up Part B For Δt = 0.5 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av. α = ∘ SubmitMy AnswersGive Up Part C For Δt = 4×10−2 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av. Express your answer using four significant figures. aav = m/s2 SubmitMy AnswersGive Up Part D For Δt = 4×10−2 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av. α = ∘ SubmitMy AnswersGive Up Part E Compare your results to the general expression for the instantaneous acceleration a⃗ for uniform circular motion that is derived in the text. Compare your results to the general expression for the instantaneous acceleration for uniform circular motion that is derived in the text. the magnitude of a⃗ av is less than of a⃗ and a⃗ av tends to a⃗ as Δt is decreased. the magnitude of a⃗ av is greater than of a⃗ and a⃗ av tends to a⃗ as Δt is decreased.

Answer 1

V^bar = 4.9 m/s. There, are two kind of acceleration: i) Tangential acceleration, a_t = dv/dt ii) Radial or centripetal acceleration: a_r = V^2/R Now, since, velocity is constant throughout so, a_t = 0 and, a_r = v^2/R = (4.9) ^2/2.9 = 8.28 m/s^2 A) Angular velocity, w = v/R = 1.69 rad/sec. so, Angle = 1.69 times 0.5 = 0.845 rad. Let, V_1 = 0i^cap + 4.9j^cap Now, Position vector at time t_2 is: X_2 = [cos(0.845) i^cap + sin(0.845) j^cap]rad so, V_2 = X_2 middot delta t = (0.332 i^cap + 0.374 j^cap) rad/sec so, a = V_2 - V_1/delta t = (0.664 i^cap - 9.052 j^cap) rad/s^2 B) Direction: theta = Tan^-1[-9.052/0.664] = 1.498 rad