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).
(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).
tension in vertical string below table = T1
m1 a = m1 g - T1
so
T1 = m1 (g-a)
tension in horizontal string above table = T2
m2 a = T2 - m2 g mu
so
T2 = m2 (a + mu g)
then
( T2 - T1 )R = I alpha
but alpha R = a
so
(T2 - T1)R = I a/R
or
T1-T2 = I a /R^2
T1 -m2(a + mu g) = I a /R^2
m1 (g-a) - m2(a + mu g) = I a /R^2
- m1 a -m2 a - Ia/R^2 = -m1 g + mu m2 g
a (m1 + m2 + I/R^2) = g (m1 - mu m2)
a = g (m1 - mu m2) / (m1 + m2 + I/R^2)
To solve this problem, we can use the following equations of motion:
1) Torque equation: τ = Ic * α
2) Linear acceleration equation: τ - T * R = (m1 + m2) * a
3) Kinetic friction equation: f_friction = μk * m2 * g
4) Tension equation: T = m1 * g - f_friction
where:
τ is the net torque acting on the pulley,
Ic is the moment of inertia of the pulley about its center of mass,
α is the angular acceleration of the pulley,
T is the tension in the string,
R is the radius of the pulley,
a is the linear acceleration of the blocks,
m1 is the mass of block 1,
m2 is the mass of block 2,
μk is the coefficient of sliding friction,
g is the acceleration due to gravity.
Now let's solve the problem:
(a) Find the magnitude of the linear acceleration of the blocks.
From equation 1:
τ = Ic * α
Since the pulley is not accelerating in the angular direction, α = 0. Therefore, the net torque acting on the pulley is also zero.
Thus, we have:
0 = Ic * α
From equation 2:
τ - T * R = (m1 + m2) * a
Since the net torque is zero, we can simplify this equation to:
0 - T * R = (m1 + m2) * a
Rearranging this equation, we get:
T = - (m1 + m2) * a / R ----(eqn 3)
From equation 4:
T = m1 * g - f_friction
Substituting the value of T from equation 3 into equation 4, we have:
- (m1 + m2) * a / R = m1 * g - μk * m2 * g
Now solving for 'a', we get:
a = - (m1 * g - μk * m2 * g) * R / (m1 + m2) ----(eqn 5)
Therefore, the magnitude of the linear acceleration of the blocks is given by:
a = - ((m1 * g) - (μk * m2 * g)) * R / (m1 + m2)
(b) How far did block 1 fall before hitting the ground?
To find the distance block 1 falls, we can use the equation of motion:
s = ut + (1/2) * a * t^2
At time t = t1, the initial velocity u of block 1 is zero, so the equation becomes:
s = (1/2) * a * t1^2
Substituting the value of 'a' from equation 5, we have:
s = (1/2) * [ - (m1 * g - μk * m2 * g) * R / (m1 + m2) ] * t1^2
Therefore, the distance block 1 falls before hitting the ground is given by:
s = (1/2) * [ - (m1 * g - μk * m2 * g) * R / (m1 + m2) ] * t1^2
To solve this problem, we can analyze the forces acting on the system. Let's denote the linear acceleration of the blocks as "a," and the tension in the string connecting the blocks as "T."
(a) To find the magnitude of the linear acceleration of the blocks, we can start by considering the forces acting on block 1 and block 2 separately.
For block 1:
- The gravitational force acting on it is m1 * g (downward).
- The tension in the string pulls it upwards with a force of T.
For block 2:
- The gravitational force acting on it is m2 * g (downward).
- The friction force opposes its motion and its magnitude is μk * m2 * g (opposite to the direction of motion).
Now, let's consider the torque about the center of the pulley caused by the net force acting on the blocks. Since the string does not slip on the pulley, the torque is equal to the moment of inertia of the pulley times its angular acceleration, which is zero because the pulley is not rotating.
The net torque about the center of mass of the pulley:
- The force from block 1 causes a torque of R * T (clockwise).
- The force from block 2 causes a torque of R * (μk * m2 * g) (counterclockwise).
Since the net torque is zero, we have R * T = R * (μk * m2 * g).
Next, considering the linear motion of the blocks, we can apply Newton's second law:
For block 1:
- The net force acting on it is T - m1 * g.
- Therefore, m1 * a = T - m1 * g.
For block 2:
- The net force acting on it is μk * m2 * g - T.
- Therefore, m2 * a = μk * m2 * g - T.
Now we have a system of equations that we can solve to find the value of "a" (the linear acceleration).
To do this, we can first eliminate "T" by substituting the value of "T" from the torque equation:
R * T = R * (μk * m2 * g)
T = μk * m2 * g
Substituting this value of "T" in the equations for block 1 and block 2, we get:
m1 * a = μk * m2 * g - m1 * g
m2 * a = μk * m2 * g - μk * m2 * g
Simplifying these equations, we have:
m1 * a = (μk * m2 - m1) * g
m2 * a = 0
Since m2 * a = 0, we can conclude that block 2 is not accelerating. Therefore, the linear acceleration of the blocks is given by:
a = (μk * m2 - m1) * g / m1
(b) To find how far block 1 falls before hitting the ground, we need to determine its distance traveled. We can use one of the kinematic equations of motion for constant acceleration:
s = ut + (1/2) * a * t^2
In this case, the initial velocity "u" is zero (as the blocks start from rest).
Using the equation with "s," we have:
s1 = 0 + (1/2) * a * t1^2
s1 = (1/2) * a * t1^2
Therefore, the distance block 1 falls before hitting the ground is given by:
s1 = (1/2) * [(μk * m2 - m1) * g / m1] * t1^2
This gives the answer in terms of m1, m2, μk, t1, and g.