Question

A 780 g , 54-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short but sharp 1000 N hammer blow at the center of the rod, aimed in a direction that causes the rod to rotate on the axle. The blow lasts a mere 2.5 ms .What is the rod's angular velocity immediately after the blow?

Answer 1

Answer:

The angular velocity is 35.5 rad/s.

Explanation:

Step 1: Find the moment of Inertia for the rod

The moment of inertia, I, of a rod:

$I=\frac{1}{12} ML^{2}$ .......... (1)

where M = mass of rod (0.78 Kg); L = Length of rod (0.54 m).

$I=\frac{1}{12} (0.78)(0.54)^{2} =0.019\frac{Kg}{m^{2} }$

Step 2: Calculate the angular acceleration from Rotational kinetic notation

F.r = I.α ........... (2)

where F is the force acting upon the rod; r is the half length of the rod; I is the moment of Inertia and; α is the angular acceleration.

∴ (1000 N)(0.27 m) = 0.019α

α = 270 Nm / 0.019 Kgm²

α = 14210.5 rad/s

Step 3: We find the angular velocity by using the equation below:

ωf = ωi + αt ......... (3)

where

ωf is the angular velocity after the blow

ωi is the angular velocity before the blow = 0

t is the time taken for the blow to occur = 2.5 ms

ωf = 0 + (14210.5 rad/s)(2.5 ms) = 35.5 rad/s.