A 500 steel block rotates on a steel table while attached to a 2.0--long massles

Question

A 500 steel block rotates on a steel table while attached to a 2.0--long massless rod. Compressed air fed through the rod is ejected from a Inozzle on the back of the block, exerting a thrust force of 3.8 . The nozzle is 70 from the radial line, as shown in the figure (Figure 1) . The block starts from rest.

here is the work i have however this online is telling me im wrong, what am i doing wrong!

The tangential thrust = T*sin(70) = 3.8N*sin(70) = 3.570N and the outward component = 3.8N*cos(70) = 1.299676N
So the torque ? = 3.570N*2.0m = 7.14N-m

The moment of inertia I = m*r^2 = 0.500kg*2^2 = 2.0kg-m^2

So the angular acceleration ? = ?/I = 7.14N-m/2.0kg-m^2 = 3.57 rad/s^2

So using the kinematic eqn ?^2 = ?o^2 + 2 ? *?..........?o = 0

so ? = sqrt(2*3.57*10rev*(2? rad/rev)) = 21.18 rad/s

Explanation / Answer

Part (A):

The tangential thrust = T*sin(70) = 3.5N*sin(70) = 3.29N and the outward component = 3.5N*cos(70) = 1.20N

So the torque ? = 3.29N*2.0m = 6.58N-m

The moment of inertia I = m*r^2 = 0.500kg*2^2 = 2.0kg-m^2

So the angular acceleration ? = ?/I = 6.58N-m/2.0kg-m^2 = 3.29 rad/s^2

So using the kinematic eqn ?^2 = ?o^2 + 2 ? *?..........?o = 0

so ? = sqrt(2*3.29*10rev*(2? rad/rev)) = 20.3 rad/s

Part (B):

We have T - outward thrust = m*r*?^2 (centripetal acceleration)

So tension T = 1.20N + 0.500kg*2.0m*(20.3)^2 = 413N

Cheers, mate!
Hope this helped. Please rate 5 if it did.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.