I have a test Monday and i'm trying some review problems my professor gave us to study from. I'm not looking for answers i'm just looking for someone to help step me through a few of these problems. Thanks for any and all help you guys can provide.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

3a). After starting, a lawn mower blade turns at a constant angular velocity of 20 rad/s. In a 10s interval, how many radians and how many degrees does it turn?

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I figured out this part. It's 200 radians (20X10) and 11459.156 degrees ( 200(180/[tex]\pi[/tex]))

b) If the blade has a radius of .3 m, what is the linear speed of the tip of the blade?

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Using v=rω, I got 6 by doing (.3)(20)= 6

c) If the blade is considered as a rectangular plate of mass .4 kg, and a width of .05 m, what is its kinetic energy as it spins.

This question (3c) is what I am stuck on.

2. Relevant equations

Moment of Inertia for a rectangular plate is: I= 1/12M(a^{2}+b^{2}) where a and b are the length and width of the rectangle.

Kinetic Energy is : K=(1/2)Iω^{2}

Edit: I dont don't why the omega symbol is there as a superscript. It should be omega squared. Sorry i'm trying to get used to this.

3. The attempt at a solution

Since I know the equations, i was going to plug the mass and dimensions into the equation I= 1/12M(a^{2}+b^{2}) to get Inertia but I realized that they only gave me one dimension (the width) and I don't know what to do since I was not given a second dimension. I'm not sure if my teacher forgot to include the second dimension or if this is actually part of the problem.

After i found Inertia i was going to use the equation K=(1/2)Iω^{2}since I knew what the Inertia and what Omega is, and use that to find the Kinetic Engery. I'm just at a standstill with the first equation.

1. The problem statement, all variables and given/known data

6. A particle of mass .4kg is attached to the 100-cm mark of a meter stick of mass .1 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50cm mark and (b) perpendicular to the table through the 0 cm mark.

ω=4 rad/s

mass of particle = .4kg

mass of meter stick =.1kg

Lengh of stick = 100cm = 1 m

2. Relevant equations

For part a) Moment of Inertia for a Long thin rod with rotation through center: I= (1/12)ML^{2}

For part b) Moment of Inertia for a Long thin rod with rotation through end: I= (1/3)ML^{2}

For these equations, L = the length of the rod and M = the total mass

To find angular momentum, use the equation L= Iω

L represents angular momentum in this equation.

3. The attempt at a solution

I know the answers. a) is .433 kgm^{2}/s and b) is 1.73 kgm^{2}/s.

But when I do the problem I don't get the right answers.

For (a), I did I= (1/12)(.5)(1^{2})

where .5 is the total mass of the stick and the particle and 1 is the length of the stick in meters. I got about .04167 and when I mutiplied that by omega (which is 4) I got .16667

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