The balance wheel of a watch is a thin ring of radius 0.95 cm and oscillates with a frequency of 3.10 Hz. If a torque of 1.1x10-5 Nm causes the wheel to rotate 45°, calculate the mass of the balance wheel.
Ok, so I set up a force diagram and did the following work but I'm stuck again...
After it's been stretched:
This then simplifies to:
I wrote α as the second derivative of θ with respect to time but now I'm stuck...
A uniform meter stick of mass M is pivoted on a hinge at one end and held horizontal by a spring with spring constant k attached at the other end. If the stick oscillates up and down slightly, what is its frequency?
Ah, you again. Thanks for the help! :)
I wish I could show you the diagram on my paper... it demonstrates the problem a lot better. The axle itself is rotating with angular velocity ω. There might not necessarily be a ball at the end of the axle but the point at the end of the axle is...
A ball of mass m is attached via a rod of length x to an axle that rotates with angular velocity ω. You can consider the ball to be a point mass.
m = 5 kg, x = 0.3 m, y = 0.4 m, ω= 30 rad/s
(a) What is the linear momentum (direction and magnitude) of the ball?
A solid disk of mass m and radius R rolls without slipping with a velocity v. Assuming it doesn't slip, how far vertically will it roll up an incline?
The Attempt at a Solution
I'm thinking that we need to...
A string passing over a pulley has a 3.80-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0 cm and mass 0.80 kg. If the bearings of they pulley were frictionless, what would be the acceleration...
Oh, okay. So the second height is ro and I get the right answer... my question is how does that make sense?
If ro is the radius of the ball but the ball is already at the bottom of the ramp, then how could the radius of the ball be part of potential energy? In order to reduce the...
Oh, okay, that makes a lot more sense. However, I got a final answer of √(10/7)(gro)
I'm thinking that I did something wrong on the left side of the equation.
The original height of the ball at the beginning would be Ro, right?
If that's the case, I end up with gRo-g(Ro-ro).
Okay, well now I'm absolutely lost. If we assume the initial position of the ball to be 0, then the final height would be Rocos(45) because 135-90=45. We're changing quadrants that the ball is in so now the 135° angle is a 45° angle.
If I'm wrong, then I have absolutely no clue where to...
I got the 45 degree angle from the 135 degree angle. If we make the center of the circle the origin of the plane, then the y component of Ro would be Rocos(45). Right?
Also, why do I need the radius of the ball? I have it as part of the moment of inertia formula for the sphere but, other...
Okay, if I make the origin of the coordinate plane the center of the circle, then h1 would be 0 and h2 would be Ro-Rocos(45°)
A ball of radius ro rolls on the inside of a track of radius Ro. If the ball starts from rest at the vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?
Included below is a link to the diagram...
An object of mass m is moved straight up by a distance h. The work that gravity does on the object is:
(e) None of the Above
Work = ∫F dl
PE = mgh
The Attempt at a Solution
I have a feeling the answer is (c)...
Hm, okay. I tried setting the total energy at the top of hill one equal to the total energy at the top of hill two.
0.5m(v1)^2+mg(h1) = 0.5m(v1)^2+mg(h2)
0.5(v1)^2 + g(h1) = 0.5(v1)^2 + g(h2)
v = sqrt(2*g*(h2-h1))
v = sqrt (2*9.81*(20-10))
v = 14 m/s
Would that work?
A roller coaster starts at the top of hill that is 10 m high. If it is to barely make it to the top of a second hill that is 20 m high, how fast must the initial speed of the roller coaster be? Assume that the roller coaster is frictionless.
Ok. I took the absolute value of what I got and ended up with an answer of 4.4 * 10^10 m/s. However, the answer is supposed to be 5220 m/s. Obviously, I'm WAY off. Bummer... once again, I'm stuck. Is my general approach correct? Thanks so much for working through this with me, I really...
Okay, I just did a lot of work and ended up with the square root of a negative number.
KE = 0.5m(v1)^2
GPE = -(GmM)/(r1)
KE = 0.5m(v2)^2
GPE = -(GmM)/(r2)
Total: 0.5m(v2)^2 - (GmM)/(r2)
0.5m(v1)^2-(GmM)/(r1) = 0.5m(v2)^2-(GmM)/(r2)...
So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).
When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)
If I plug that in to the centripetal force equation, I get
F = (mv^2)/r
F = (Nv^2)/((r)(a+g))
Is this right? If so, where do I...