A 0.17 kg ball on a 0.70 m string rotates in a circular path in a vertical plane. If the angular speed of the ball at the bottom
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Answer:
v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²
Explanation:
The expression left corresponds to an oscillatory movement (MAS), the speed is defined by
v = dr / dt
the function of position
r = 2 cos πt i^ + 3 sin πt j^
let us note that it is a movement in two dimensions
let's perform the derivative
v = -2π sin πt i^ + 3π cos πt j^
we evaluate this expression for t = 0.25 s, remember that the angle is in radians
v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^
v = (-4.44 i^ + 6.66 j^ ) m/s
To calculate the mean acceleration we use the expression
a = () / Δt
indicates that the time is the first 3 s
we look for the initial velocity t = 0 s
v₀ = 0 i ^ + 3π j ^
we look for the fine velocity, t = 3 s
v_f = - 2π sin (π 3) + 3π cos (π 3) j ^
v_f = 0 i ^ - 3π j ^
we calculate the average acceleration
Δt = (3 -0) = 3 s
a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3
a_average = (0 i ^ -2π j ^ ) m/s²