An Atwood's machine consists of blocks of masses m1 = 13.0 kg and m2 = 22.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 6.60 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.(a) Why must the tension T2 be greater than the tension T1? (b) What is the acceleration of the system, assuming the pulley axis is frictionless? (Give the magnitude of a.)____m/s2(c) Find the tensions T1 and T2.T1 =_____NT2 =_____N The image illustrates an Atwood machine setup, which consists of a single pulley with two masses, \( m_1 \) and \( m_2 \), suspended on either side.Components:1. **Pulley**: - Represented by a circular disk with radius \( r \) and mass \( M \). - The pulley is mounted on an axle, which allows it to rotate.2. **Masses**: - \( m_1 \) is the mass on the left, depicted as a spherical object. - \( m_2 \) is the mass on the right, shown as a cuboidal object.3. **Tensions**: - \( T_1 \) denotes the tension in the rope connected to \( m_1 \). - \( T_2 \) represents the tension in the rope attached to \( m_2 \).4. **Acceleration**: - The diagram shows acceleration vectors \( \vec{a} \) indicating the direction of acceleration for each mass. - \( m_1 \) accelerates upwards while \( m_2 \) accelerates downwards.This setup is commonly used in physics to study the mechanics of rotational systems and the effect of different forces on motion. The differences in mass (\( m_1 \) and \( m_2 \)) cause the system to accelerate, leading to rotational motion of the pulley.

m1 = 13 kg m2 = 22 kg M = 6.6 Kg r = 0.2 m a) T2 must be greater than T1 because then only the torque exerted by T2 would be greater and the mass m2 would fall in the downwards direction otherwise this mass would start rising upwards and the other mass would start to descend downwards. b)For T2 = m2 ( g -a)T2 = 22 (g -a) ....(1)For T1 = m1 ( g +a)T1 = 13 (g +a) ....(2)Now , torque experienced by the pulley : (T2 - T1) * r = (1/2)M(r)2*(a/r)T2 - T1 = 6.6a/2Put value from equation (1) and (2) 22 (g -a) - 13 (g +a) = 3.3 a22g - 22a - 13g - 13a = 3.3 a 9 g = 38.3 aa = 2.303 m/s2c)Put value of a in equation 1 and 2 T2 = 22 ( 9.8 - 2.3028)T2 = 164.94 N T1 = 13 (9.8 + 2.3028)T1 = 157.3 N