Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Physics HELP regarding Rotational Kinematics?
1) A wheel with a radius of 0.7m and a moment of inertia 8kg m2 has an initial rotational velocity of 7 rad/s. A person takes a block and presses it against the edge of wheel, slowing it down with friction. The coefficient of kinetic friction between the wheel and the block is 0.4 and the person presses with a force that increases as a function of time: (8t) N.
How long does it take for the wheel to stop?
How far did the wheel rotate in the time it took to stop?
2) A block of mass 9kg is moving in a circular path on a tabletop. The radius of the circle is 0.5m and the object's speed is 4m/s.
What is the initial angular momentum of the system? kg m^2/s
What is the initial kinetic energy of the system? J
Suppose the mass was being pulled in circular motion by a string. The string is threaded through a small hole on the top of the table, and a person pulls on the string until it has a length of 0.4m.
What is the new angular momentum of the system? kg m^2/s
What is the new speed of the mass? m/s
What is the new kinetic energy of the system? J
Comparing kinetic energies, how much work was done? J
On paper, find the work done on the object by integrating the work done as the mass is pulled inwards. The force in question will be the tension in the string, which is known by calculating the centripetal force as a function of string length. Note that velocity is a function of string length
1 Answer
- WoodsmanLv 77 years agoFavorite Answer
1) I don't recall seeing a problem like this one before, so I'll try it.
τ = dL/dt = d(Iω)/dt = 8kg·m² * dω/dt
But also τ = r x F which in this problem means τ = -0.7m * 0.4 * 8N/s * t
or τ = -2.24N·m/s * t where t is in seconds. Since τ = τ, we have
8kg·m² * dω/dt = -2.24N·m/s * t
dω = -0.28rad/s³ * t * dt → integrate both sides
ω = -0.14rad/s³ * t² + C
When t = 0, ω = 7 rad/s, so C = 7 rad/s
When ω = 0, what is t ?
0.14rad/s³ * t² = 7 rad/s
t = 50 s²
t = 7.1 s ← time to stop
For the distance, I'll make a (reasonable ?) guess that
Θ = ∫ ω(t) dt = ∫ (-0.14rad/s³ * t³ + 7rad/s) dt
Θ = -0.035rad/s³ * t^4 + 7rad/s * t → for 0 ≤ t ≤ 7.1 s
Θ = -0.035rad/s³ * (7.1s)^4 + 7rad/s * 7.1s = -38 rads
OK, so I've got the wrong sign; but the magnitude looks reasonable.
That's about 6 revs.
