A pulley of mass mp , radius R , and moment of inertia about the center of mass Ic=12mpR2 , is suspended from a ceiling. The pulley rotates about a frictionless axle. An inextensible string of negligible mass is wrapped around the pulley and it does not slip on the pulley. The string is attached on one end to an object of mass m1 and on the other end to an object of mass with m2<m1 .
At time t=0 , the objects are released from rest.
(a) Find the magnitude of the acceleration of the two objects. Express your answer in terms of m1, m2, mp, R and acceleration due to gravity g (enter m_1 for m1, m_2 for m2, m_p for mp, R for R and g for g).
a=
(b) How long does it take the objects to move a distance d? Express your answer in terms of m1, m2, mp, d and acceleration due to gravity g (enter m_1 for m1, m_2 for m2, m_p for mp, d for d and g for g).
t=
a=((m_1-m_2)*g)/(((0.5*m_p)+(m_1+m_2)))
t=sqrt((d*(m_p+2*(m_1+m_2)))/((m_1-m_2)*g))
To find the magnitude of the acceleration of the two objects, we need to analyze the forces acting on them and apply Newton's second law. Let's consider each object separately:
For object with mass m1:
1. The weight of the object is mg, where g is the acceleration due to gravity.
2. The tension in the string is T.
3. The net force acting on m1 is T - mg.
For object with mass m2:
1. The weight of the object is m2g.
2. The tension in the string is T.
3. The net force acting on m2 is T - m2g.
Now, let's consider the torque equation for the pulley:
τ = Iα
Since the pulley rotates about a frictionless axle, the torque τ is given by:
τ = TR, where T is the tension in the string and R is the radius of the pulley.
The moment of inertia of the pulley about its center of mass is Ic = 12mpR^2.
Now, we can write the net torque acting on the pulley as:
TR = Icα
Since α is the angular acceleration, which is the same for both sides of the pulley, we can rewrite it as α = a/R, where a is the linear acceleration of the objects.
Substituting the values, we have:
TR = (12mpR^2)a/R
T = (12mpR)a
Now, we can equate the net forces acting on m1 and m2 to the tensions in the string:
T - mg = ma (for m1)
T - m2g = -ma (opposite direction for m2)
Solving these equations simultaneously, we get:
T = (m1 - m2)g / (m1 + m2)
Substituting this value of T in the earlier equation, we get:
(m1 - m2)g / (m1 + m2) = (12mpR)a
a = (m1 - m2)g / (12mpR + m1 + m2)
Therefore, the magnitude of the acceleration of the two objects is:
a = (m1 - m2)g / (12mpR + m1 + m2)
Now, let's move on to part (b) and find the time it takes for the objects to move a distance d.
We know that the distance traveled by an object with constant acceleration a is given by:
d = 0.5at^2
Rearranging the equation, we get:
t = sqrt((2d) / a)
Substituting the value of a we found earlier, we get:
t = sqrt((2d) / ((m1 - m2)g / (12mpR + m1 + m2)))
Therefore, the time it takes for the objects to move a distance d is:
t = sqrt((24dmpR + 2d(m1 + m2)) / ((m1 - m2)g))