A wheel in the shape of a uniform disk of radius R and mass mp is mounted on a frictionless horizontal axis. The wheel has moment of inertia about the center of mass Icm=(1/2)mpR2 . A massless cord is wrapped around the wheel and one end of the cord is attached to an object of mass m2 that can slide up or down a frictionless inclined plane. The other end of the cord is attached to a second object of mass m1 that hangs over the edge of the inclined plane. The plane is inclined from the horizontal by an angle θ . Once the objects are released from rest, the cord moves without slipping around the disk. Find the magnitude of accelerations of each object, and the magnitude of tensions in the string on either side of the pulley. Assume that the cord doesn't stretch (a1=a2=a). Express your answers in terms of the masses m1, m2, mp, angle θ and the gravitational acceleration due to gravity near earth's surface g (enter m_1 for m1, m_2 for m2, m_p for mp, theta for θ and g for g).
a1=a2=a=
T1= (where the string is connected to m1)
T2=(where the string is connected to m2)
http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/two-dimensional-rotational-motion/two-dimensional-rotational-dynamics/MIT8_01SC_problems21_soln.pdf
Its a different question.. please help
To find the magnitude of accelerations of each object and the magnitude of tensions in the string on either side of the pulley, we can use the principles of Newton's laws and torque.
Let's start by considering the forces acting on each object individually:
1. Object m1 (hanging over the edge of the inclined plane):
- There is only one force acting on m1, which is its weight mg pulling it downwards.
- The tension in the string connected to m1 is directed upwards.
Using Newton's second law (F = ma) for m1 in the vertical direction:
mg - T1 = m1 * a1
2. Object m2 (sliding up or down the inclined plane):
- There are two forces acting on m2. Its weight mg is directed downwards, and there is a component of the tension in the string acting upwards along the inclined plane.
- The tension in the string connected to m2 is directed downwards.
Using Newton's second law for m2 in the vertical direction:
T2 - mg * sin(θ) = m2 * a2
Using Newton's second law for m2 in the horizontal direction:
mg * cos(θ) = m2 * a2
Now let's consider the rotational motion of the wheel (uniform disk) with moment of inertia Icm:
- The net torque on the disk is given by the difference between the torque due to the tension on one side of the pulley (T1 side) and the torque due to the tension on the other side (T2 side).
- The torque due to the tension on one side can be calculated as T1 * R (where R is the radius of the wheel), while the torque due to the tension on the other side is T2 * -R.
- The net torque is equal to the moment of inertia Icm multiplied by the angular acceleration.
Using the torque equation:
T1 * R - T2 * R = Icm * α
Since the cord moves without slipping around the disk, we have a1 = a2 = a. Also, we can relate the angular acceleration α to the linear acceleration a using α = a/R.
Simplifying the equations, we can rewrite them as follows:
1. mg - T1 = m1 * a
2. T2 - mg * sin(θ) = m2 * a
3. mg * cos(θ) = m2 * a
4. T1 - T2 = (1/2)mp * R^2 * (a/R)
To solve for the acceleration a, we can substitute equations 3 and 4 into equation 2:
(mg * cos(θ)) - (1/2)mp * R^2 * (a/R) - mg * sin(θ) = m2 * a
Simplifying and rearranging the terms, we get:
(m1 + m2 + (1/2)mp) * a = mg * (cos(θ) - sin(θ))
Finally, we can solve for a:
a = (mg * (cos(θ) - sin(θ))) / (m1 + m2 + (1/2)mp)
To find the tensions in the string on either side of the pulley, we can substitute the value of a into equations 1 and 2:
T1 = mg - m1 * a
T2 = mg * sin(θ) + m2 * a
So, the solutions are:
a1 = a2 = a = (mg * (cos(θ) - sin(θ))) / (m1 + m2 + (1/2)mp)
T1 = mg - m1 * a
T2 = mg * sin(θ) + m2 * a