Recent content by qqchan

Here is the image. I decided to accomplish it this way, but it still didn't work. v = (mball*vball + mtall*0.13*ω)/(mball + mtall) I believe I have made it all to the same dimensions

yes, that is right. and I is the moment of inertia of the object about its centre. I could calculate I by using the conservation of linear momentum. However, I do not know what they mean by centre of mass velocity.

Homework Statement As part of a carnival game, a 0.523-kg ball is thrown at a stack of 19.5-cm tall, 0.227-kg objects and hits with a perfectly horizontal velocity of 9.9 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision...

Hmm, that is right

τnet = 0; FRsinθ = mgRsin(90-θ) F/(mg)=cotθ (mg)/F = tanθ Solve for θ since sinθ = 1-n. This way we could solve for n. I guess this is the way

Iw2 would be the radial. Iw2 = F τnet = -Frsinθ + mgR/2sin(90-θ) I guess these are the formulas

if Fn = normal force. So, if it is tangential to the wheel, the best way to make it work would be the torque. There is no frictional force in this case. τ=Frsinθ F = Fn in such case. is this what you are suggesting?

So after changing the direction of the normal force. Here are the force vectors I get: F = Fn Fg = ma τnet = 0 τnet = -Frsinθ + mgR/2sinθ Fr = mgR/2sinθ F = mg/2sinθ 2F/mg = sinθ 2F/mg = 1-n Am I missing something? Or did I do write an equation wrong.

So according to your diagram tanθ=Mg/2F The torque where the points meet. The formula for torque is τ=Frsinθ

Haruspex, I have decided to remove my thought on torque and drew the body diagram again. sin θ = (R-h)/R θ = 1-h/R θ = 1-n Is this what you are suggesting?

Theta is the angle between the radius and the point of contact with the step. However, I am confused with "the angle that the radius from wheel centre to point of contact with step makes to the horizontal." I got mgR/2 from the thought of finding the center of mass

This is the image I have came out with. From this picture I can conclude that τ = Frsinθ The known values are: F = 410 I know that Force of gravity is pushing down and the torque for wheel pushing upward. Hence, mgR/2 = FRsinθ (mg)/(2F) = sinθ Is the approach correct? because I could not...

Homework Statement A person pushing horizontally a uniformly-loaded 45.2 kg wheelbarrow is attempting to get it over a horizontal step. The maximum horizontal force that person could apply is F = 410 N. What is the maximum height of the step, as a fraction of the wheel's radius, the person...

Homework Statement A newly discovered planet has a mean radius of 1230 km. A vehicle on the planet's surface is moving in the same direction as the planet's rotation, and its speedometer reads 130 km/h. If the angular velocity of the vehicle about the planet's center is 6.18 times as large...