Recent content by tachu101

Friction= (1/2)(M)(a(linear)) ? then I plug that into solve for the values?

Net torque= Friction(disk)* Radius= Ia(rotational)= (1/2)(M)(R^2)(alinear/R)? Is this then ((mgcostheta)*Uk)*R= (1/2)(M)(R^2)(a linear/R). This will solve for the linear acceleration of the disk. And that is the same as the linear acceleration of the mass right? For the friction force acting...

Homework Statement ln(X+1)=Sin^2(x) solve The Attempt at a Solution I can get three solutions 0, .964, 1.684 using the calculator. How do you solve this wo/ a calculator.

For Disk : Fx= Wsintheta-friction-T=Ma For Mass : Fx= Wsintheta-friction+T=Ma For both: Fy= N-Wcostheta=0 Now What?

\sumFx= Wsin\Theta(disk)-Friction(disk)-Tension+Tension+Wsin\Theta(mass)-Friction(mass)= Ma \sumFy= N-Wcos\Theta=0

Homework Statement A dis of mass (m) and radius (r) with I=(1/2)(m)(r^2) is rolling down an incline dragging with it a mass (m) attached with a light rod to a bearing at the center of the disc. The friction coefficients are the same for both masses, Uk and Us. Determine the linear...

a=torque/I == (F*(L/4))/ (2(m(L/2)^2)) = a= F/2mL ? a(linear)= a(rot)*radius == F/2ml= a*(L/2) solve for a and that's the answer?

I'm really lost at what to do in this question.

torque = r*F... torque= (L/4)(F) right?

Homework Statement Two equal masses (m) are connected by a light rod of length L that is pivoted about its center. A downward Force F is applied to the rod at a distance L/4 from the pivot. If you ignore gravity, the linear acceleration of the mass is... O--------o--------OHomework Equations...

pi/2 would be the angle needed to close the gate, but 1.35 is less than pi/2, so it would not be closed in time.

If I use torque=Ia and solve for a I will get a=torque/I = (300)/(1000) = .3 rad/sec^2 as the angular acceleration. For the w value after 3 sec it would be .9rad/sec? So I could then do the KE= (1/2)(I)(w^2) = (1/2)(1000)(.9^2) = 405 Then for part c could I use Theta= at^2/2 and get...

I looked back and I think I missed a 1/2 in part a. It should be (1/2)((1/2)(1)(2^2))(3^2)= 9 I think I missed the 1/2 in the I value.

Homework Statement A gate to a bullpen is open at a right angle to the fence and the bull is rushing toward the opening to get out. The farmer estimates that the bull will reach the opening in 3 sec. so he pushes at the end of the gate (always at a right angle) with a force of 100N. The moment...

In summary... I think that part a is 9J because I think it is (1/2)(I)(w^2)= 9 Part b is then torque*W which is then (10)(2)*(3)= 60