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=
m₁a=m₁g-T₁
m₂a=T₂- F(fr)
0=m₂g-N
Iε=(T₁-T₂)R =>T₁-T₂ = Iε/R=Ia/R²
N= m₂g,
F(fr)=μN=μm₂g
m₁a+ m₂a= m₁g-T₁+ T₂- F(fr)=
=m₁g- μm₂g –( T₁-T₂) =
=m₁g- μm₂g – Ia/R².
a=g(m₁-μm₂)/{m₁+ m₂+(I/R²)}.
d=at²/2
To find the linear acceleration of the blocks, we need to analyze the forces acting on the system.
First, let's consider Block 1. The only force acting on it is its weight, which is given by m1 * g, directed vertically downwards.
Now, let's consider Block 2. There are two forces acting on it: the tension in the string (T), directed horizontally, and the friction force (f), directed opposite to the direction of motion.
Applying Newton's second law to Block 1, we have:
m1 * g = m1 * a
where a is the linear acceleration of the blocks.
Now, let's consider the forces acting on Block 2. The tension in the string causes an equal and opposite force on Block 2. So, we have:
T - f = m2 * a
The tension in the string is also equal to the torque exerted by the pulley. The torque can be calculated as the product of the radius, R, and the tension, T:
T = Ic * α / R
where α is the angular acceleration of the pulley.
The friction force, f, can be calculated as the product of the coefficient of friction, μk, and the normal force, which is equal to m2 * g. So, we have:
f = μk * m2 * g
Substituting the expressions for T and f into the equation for Block 2, we get:
Ic * α / R - μk * m2 * g = m2 * a
Now, we need to relate the linear acceleration, a, to the angular acceleration, α. The pulley has a radius, R, so the linear acceleration of the blocks at the edge of the pulley is equal to the angular acceleration multiplied by the radius:
a = α * R
Substituting this expression into our equation, we get:
Ic * α / R - μk * m2 * g = m2 * α * R
Simplifying the equation, we have:
Ic * α - μk * m2 * g * R = m2 * α * R^2
Dividing both sides of the equation by R^2, we get:
Ic * α / R^2 - μk * m2 * g = m2 * α
Now, we can solve for α:
α = μk * m2 * g / (Ic / R^2 + m2)
Substituting the expression for α into the equation for a, we get:
a = α * R = (μk * m2 * g / (Ic / R^2 + m2)) * R
Simplifying the expression, we have:
a = μk * m2 * g * R / (Ic + m2 * R^2)
Therefore, the magnitude of the linear acceleration of the blocks is:
a = (μk * m2 * g * R) / (Ic + m2 * R^2)
To find how far Block 1 falls before hitting the ground, we can use the kinematic equation:
d = (1/2) * g * t^2
where d is the distance, g is the acceleration due to gravity, and t is the time.
Since Block 1 hits the ground at time t1, the distance it traveled is:
d = (1/2) * g * t1^2
Therefore, the distance Block 1 falls before hitting the ground is:
d = (1/2) * g * t1^2