Question

. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the arm is fully extended the speed of the hand uniformly increases from 1 m/s to 2.5 m/s in 3 seconds, determine the magnitude of the acceleration at the instant the speed of the hand is 2.0 m/s.

Answer 1

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

2.5 = 1 + 3 a

a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

$A=\sqrt{a^{2}+a'^{2}}$

$A=\sqrt{0.5^{2}+13.3^{2}}$

A = 13.309 m/s²