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)

m₁a=m₁g-T₁

m₂a=T₂ - m₂gsinα
Iε=(T₁-T₂)R =>
T₁-T₂ = Iε/R=Ia/R²=mR²a/2 R²=ma/2

m₁a+ m₂a= m₁g-T₁+ T₂ -
=m₁g- m₂gsinα –( T₁-T₂) =
=m₁g- m₂gsinα – ma/2
a(m₁+ m₂+0.5m)=g (m₁- m₂sinα)
a= g (m₁- m₂sinα)/ (m₁+ m₂+0.5m)
T₁=m₁(g-a)
T₂=m₂(a+gsinα)

Please give correct answer for this

To find the magnitudes of accelerations of each object and the magnitudes of tensions in the string on either side of the pulley, we can apply Newton's laws of motion.

First, let's consider the object of mass m1 hanging over the edge of the inclined plane.

1. Find the acceleration of mass m1:
- The net force acting on mass m1 is the tension in the string (T1) minus the gravitational force (m1 * g * cos(θ)).
- According to Newton's second law, the net force is equal to the mass (m1) multiplied by the acceleration (a1).
- Set up the equation: T1 - m1 * g * cos(θ) = m1 * a1.

2. Find the acceleration of mass m2 sliding on the inclined plane:
- The net force acting on mass m2 is the tension in the string (T2) minus the gravitational force (m2 * g * sin(θ)).
- According to Newton's second law, the net force is equal to the mass (m2) multiplied by the acceleration (a2).
- Set up the equation: T2 - m2 * g * sin(θ) = m2 * a2.

3. Use the equation for rotational motion:
- The tension in the string (T2) also causes a torque that causes the wheel to rotate. The torque is equal to the moment of inertia (Icm) multiplied by the angular acceleration (α).
- The angular acceleration (α) is related to the linear acceleration (a) and the radius (R) by the equation α = a / R.
- Therefore, the torque can be expressed as T2 * R = Icm * (a / R).
- Simplify the equation: T2 = (1/2) * mp * R * a.

4. Solve the system of equations:
- Substitute the expression for T2 from step 3 into the equation from step 2: (1/2) * mp * R * a - m2 * g * sin(θ) = m2 * a2.
- Substitute the expression for a2 from step 4 into the equation from step 1: T1 - m1 * g * cos(θ) = m1 * (1/2) * mp * R * a - m2 * g * sin(θ).

Now, you can solve these equations simultaneously to find the values of a1, a2, T1, and T2 in terms of the given variables m1, m2, mp, θ, and g.