A pulley of mass mp , radius R , and moment of inertia about its center of mass Ic , is attached to the edge of a table. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to block 1 that hangs over the edge of the table. The other end of the string is attached to block 2 which slides along a table. The coefficient of sliding friction between the table and the block 2 is μk . Block 1 has mass m1 and block 2 has mass m2 , with m1>μkm2 . At time t=0 , the blocks are released from rest. At time t=t1 , block 1 hits the ground. Let g denote the gravitational acceleration near the surface of the earth.
(a) Find the magnitude of the linear acceleration of the blocks. Express your answer in terms of m1 , m2, Ic, R, μk and g as needed (enter m_1 for m1, m_2 for m2, I_c for Ic, R for R, mu_k for μk and g for g).
a=
(b) How far did the block 1 fall before hitting the ground? Express your answer in terms of m1 , m2, Ic, R, μk, t1 and g as needed (enter m_1 for m1, m_2 for m2, I_c for Ic, R for R, mu_k for μk, t_1 for t1 and g for g).
d=
g*(m_1-mu_k*m_2)/(m_1+ m_2+(I_c/R^2))
g*(m_1-mu_k*m_2)/(m_1+ m_2+(I_c/R^2))*t_1^2/2
Thanx a lot
To find the magnitude of the linear acceleration of the blocks, we need to consider the forces acting on the system.
Let's analyze the forces on block 1:
1. Gravitational force: The weight of block 1 acts vertically downward and has a magnitude of m1 * g.
2. Tension force: The tension in the string acts upwards and is equal to the tension in the string on the other side. Therefore, we have Tension1 = Tension2.
Now let's analyze the forces on block 2:
1. Gravitational force: The weight of block 2 acts vertically downward and has a magnitude of m2 * g.
2. Frictional force: The sliding friction between block 2 and the table opposes the motion and acts in the opposite direction of the motion. Its magnitude is given by the equation frictional force = μk * Normal force, where the normal force is equal to the weight of block 2 in this case.
Now, considering the moment of inertia of the pulley and applying Newton's second law, we can write the equations of motion:
For Block 1:
m1 * g - Tension1 = m1 * a
For Block 2:
Tension2 - m2 * g - frictional force = m2 * a
We also know that the acceleration of the pulley is equal to the linear acceleration of the blocks, so we have:
a = α * R, where α is the angular acceleration of the pulley.
For the pulley, the torque acting on it is caused by the frictional force, and it is given by:
Torque = Ic * α, where Ic is the moment of inertia of the pulley.
Since the frictional force is equal to the torque divided by the radius of the pulley, we can write:
frictional force = Torque / R = (Ic * α) / R
Substituting this expression for the frictional force into the equation for Block 2, we get:
Tension2 - m2 * g - (Ic * α) / R = m2 * a
Now, we have a system of equations that we can solve for the linear acceleration, a.
To solve this system of equations, we eliminate the tensions by substituting Tension1 = Tension2:
m1 * g - m2 * g - (Ic * α) / R = (m1 + m2) * a
Next, we can substitute α = a / R into the equation:
m1 * g - m2 * g - (Ic * a) / R^2 = (m1 + m2) * a
Now, we can solve for a:
(m1 - m2) * g - (Ic * a) / R^2 = (m1 + m2) * a
(m1 - m2) * g = ((m1 + m2) * a) + (Ic * a) / R^2
(m1 - m2) * g = a * (m1 + m2 + Ic / R^2)
a = (m1 - m2) * g / (m1 + m2 + Ic / R^2)
Therefore, the magnitude of the linear acceleration of the blocks is:
a = (m1 - m2) * g / (m1 + m2 + Ic / R^2)
For part (a), the magnitude of the linear acceleration of the blocks is:
a = (m1 - m2) * g / (m1 + m2 + Ic / R^2)
To find the distance block 1 falls before hitting the ground, we need to find its displacement. Since the blocks were initially at rest and block 1 hits the ground at time t = t1, we can use the equation of motion:
d = (1/2) * a * t1^2
Substituting the expression for a from part (a), we get:
d = (1/2) * [(m1 - m2) * g / (m1 + m2 + Ic / R^2)] * t1^2
Therefore, the distance block 1 falls before hitting the ground is:
d = (1/2) * [(m1 - m2) * g / (m1 + m2 + Ic / R^2)] * t1^2
For part (b), the distance block 1 falls before hitting the ground is:
d = (1/2) * [(m1 - m2) * g / (m1 + m2 + Ic / R^2)] * t1^2