An object of mass m1 = 4.90 kg is connected by a light cord to an object of mass m2 = 3.00 kg on a frictionless surface (see figure). The pulley rotates about a frictionless axle and has a moment of inertia of 0.440 kg · m2 and a radius of 0.320 m. Assume that the cord does not slip on the pulley.An illustration shows a rectangular block of mass m2 on the horizontal surface of a table. Block m2 is connected to another rectangular block of mass m1 by a cord that runs over a pulley placed diagonally at the corner of the horizontal surface of the table. The rectangular block of mass m1 is suspended vertically by the side of the tabletop. The part of the cord that extends from m1 to the pulley is labeled T1, and the part of the cord that extends from m2 to the pulley is labeled T2.(a) Find the acceleration of the two masses. m/s2(b) Find the tensions T1 and T2.T1 = NT2 = N
Question
An object of mass m1 = 4.90 kg is connected by a light cord to an object of mass m2 = 3.00 kg on a frictionless surface (see figure). The pulley rotates about a frictionless axle and has a moment of inertia of 0.440 kg · m2 and a radius of 0.320 m. Assume that the cord does not slip on the pulley.An illustration shows a rectangular block of mass m2 on the horizontal surface of a table. Block m2 is connected to another rectangular block of mass m1 by a cord that runs over a pulley placed diagonally at the corner of the horizontal surface of the table. The rectangular block of mass m1 is suspended vertically by the side of the tabletop. The part of the cord that extends from m1 to the pulley is labeled T1, and the part of the cord that extends from m2 to the pulley is labeled T2.(a) Find the acceleration of the two masses. m/s2(b) Find the tensions T1 and T2.T1 = NT2 = N
Solution
To solve this problem, we need to use Newton's second law (F = ma) and the rotational equivalent (τ = Iα), where F is the net force, m is the mass, a is the acceleration, τ is the net torque, I is the moment of inertia, and α is the angular acceleration.
(a) Find the acceleration of the two masses.
First, we need to find the net force acting on the system. The force due to gravity on m1 is m1g (where g is the acceleration due to gravity, approximately 9.8 m/s²), and the force due to gravity on m2 is m2g. The net force is then m1g - m2g.
Next, we need to find the net torque acting on the pulley. The tension in the cord causes a torque of T1r (where r is the radius of the pulley), and the weight of m1 causes a torque of m1gr. The net torque is then T1r - m1gr.
Setting the net force equal to ma (where a is the acceleration of the masses) and the net torque equal to Iα (where α is the angular acceleration, which is equal to a/r), we can solve for a:
m1g - m2g = (m1 + m2 + I/r²)*a
Solving for a gives:
a = (m1g - m2g) / (m1 + m2 + I/r²)
Substituting the given values gives:
a = (4.90 kg * 9.8 m/s² - 3.00 kg * 9.8 m/s²) / (4.90 kg + 3.00 kg + 0.440 kg*m² / (0.320 m)²) a = 2.45 m/s²
(b) Find the tensions T1 and T2.
The tension T1 is the force required to accelerate m1 upward, which is m1a + m1g. Substituting the given values gives:
T1 = 4.90 kg * 2.45 m/s² + 4.90 kg * 9.8 m/s² T1 = 60.1 N
The tension T2 is the force required to accelerate m2 horizontally, which is m2*a. Substituting the given values gives:
T2 = 3.00 kg * 2.45 m/s² T2 = 7.35 N
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