Bob starts spinning his hammer at a rate of 2.00 rad/s^2. At the same time, he starts running towards Jim at 4.00 m/s^2. The distance from the end of the hammer to Bob's elbow is 1.20 m. If Bob takes 3 seconds before he launches the hammer at Jim, what is the resultant velocity of the hammer when it leaves Bob's hand?
My attempt.
tangential acceleration = (1.20m)(2.00rad/s^2) = 2.4 m/s^2
resultant/total acceleration = 2.4 m/s^2 + 4.00 m/s^2 = 6.4 m/s^2
change in velocity = (6.4 m/s^2)(3.00s) = 19.2 m/s
To determine the resultant velocity of the hammer when it leaves Bob's hand, we can use the concept of kinematic equations. Let's break down the problem step by step:
1. First, let's find the change in velocity of the hammer.
a. The tangential acceleration of the hammer is given as 2.00 rad/s^2. To find the linear acceleration, we need to multiply this by the distance from the end of the hammer to Bob's elbow, which is 1.20 m.
Tangential acceleration = (1.20 m)(2.00 rad/s^2) = 2.4 m/s^2.
b. Bob is also running towards Jim at an acceleration of 4.00 m/s^2.
Total acceleration = tangential acceleration + linear acceleration = 2.4 m/s^2 + 4.00 m/s^2 = 6.4 m/s^2.
c. We are given that Bob takes 3 seconds before launching the hammer, so we can multiply the total acceleration by the time to get the change in velocity:
Change in velocity = (6.4 m/s^2)(3.00 s) = 19.2 m/s.
2. Finally, the resultant velocity is equal to the initial velocity plus the change in velocity.
Since Bob is initially at rest, the initial velocity is 0 m/s.
Resultant velocity = Initial velocity + Change in velocity = 0 m/s + 19.2 m/s = 19.2 m/s.
Therefore, the resultant velocity of the hammer when it leaves Bob's hand is 19.2 m/s.