Circular motion of steel block

[nat]

I've been trying to figure this out forever

A 500g steel block rotates on a steel table while attached to a 2.0 meter long massless rod. compressed air is fed through the rod is ejected from a nozzle on the back of the block, exerting a force of 3.5N. The nozzle is 70 degrees from the radial line. The block starts out at rest.

what is the blocks angular velocity after 10 revolutions?
what is the tesion in the rod after 10 revolutions?

I know mu kentic of steel on steel is .6 I've been running myself around in circles for this problem!

I know I have to add all the vector components together but I'm new at this stuff I can identify the force of... gravity*mass thrust(but since it is connected at an angle I don't know how eactly to divide it into components), friction, I have no idea what the acceleration is.
I can do the circular motion that has a horizontal and verticle component, but the angle thing is throwing me!

 

[rayveldkamp]

I haven't done a unit in dynamics since last semester so I am a bit rusty...but i remember getting a problem similar to that, where i used the normal-tangential coordinate system for the block's acceleration and velocity (looking down on the table), and then analysed the friction and gravitational force with x-y coordinates (looking side on at the table).
Don't know if that helps, as i said its been a while.

 

[wais]

First of all, don't worry too much about feeling stuck on this problem. Circular motion can be tricky and it's normal to struggle with it at first. Let's break down the problem step by step to help you understand it better.

First, let's start with the block's angular velocity. Angular velocity is defined as the rate of change of angular displacement over time. In simpler terms, it is how fast an object is rotating around a fixed point. In this case, the block is attached to a rod and rotating around the center of the table. Since the block starts at rest and we are given the force and distance, we can use the equation for angular acceleration:

α = τ/I

Where α is the angular acceleration, τ is the torque (force x distance), and I is the moment of inertia (a measure of how difficult it is to rotate an object). In this case, the moment of inertia can be calculated as I = mR^2, where m is the mass of the block and R is the length of the rod. Plugging in the values, we get:

α = (3.5N)(2.0m)/((0.5kg)(2.0m^2)) = 7 rad/s^2

Now, we can use the equation for angular velocity to find the block's angular velocity after 10 revolutions:

ω = ω0 + αt

Where ω0 is the initial angular velocity (which is 0 since the block starts at rest), α is the angular acceleration we just calculated, and t is the time. Since we are given that the block completes 10 revolutions, we know that t = 10T, where T is the time it takes for one revolution. Therefore:

ω = 0 + (7 rad/s^2)(10T) = 70T rad/s

To find the time it takes for one revolution, we can use the equation for angular velocity again:

ω = Δθ/Δt

Where Δθ is the change in angular displacement and Δt is the time it takes to complete that change. In this case, we know that the block goes through 2π radians (one full revolution) in T seconds, so:

T = 2π/ω = 2π/70T = 0.09 seconds

Therefore, the block's angular velocity after 10 revolutions is 70T rad/s = 70(0