Question

A pulley of moment of inertia 2.5 kg .m2 is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. Assume the following data: 11 = 49 cm, 12 = 20 cm, m = 1.0 kg, and m2 = 2.5 kg. (a) What is the angular acceleration of the pulley? (Enter the magnitude in rad/s2.) -039 X rad/s2 (b) What is the linear acceleration in m/s2) of the weights? (Indicate the direction with the signs of your answers. Assume the upward direction is positive.) a = .019 X m/s2 az = -0078 1 x m/s²

Please be the one to solve this problem no one else can the answers above are not the right answers.

Answer 1

Solution -

• Moment of inertia  = 2.5 kg-m_²
• r_{1 = 49 cm}
• r_{2 = 20 cm}
• m_{1 = 1.0 Kg}
• m_{2 = 2.5 Kg}

Use the equation - >

Net torque acting on the system = Moment of inertia x angular acceleration

T₂- T₁= Ia^, ( a^, = angular acceleration , T = Torque )

T₁ = m₁gr_{1 = 1 x 9.81 x 49 x 10^-2}    

= 4.8069

T₂ = m₂gr_{2 = 2.5 x 9.81 x 20 x 10^-2}

= 4.905

Tnet = T₂- T₁

Tnet = 4.905 - 4.8069 = 0.0981

Put value in T₂- T₁= Ia^,

0.0981 =  2.5 x  a^,

a^,   = 0.0981 / 2.5 = 0.03924 rad / sec² ( Clockwise )

Note - In answer only enter the above value because they asked for magnitude not the direction of rotation .

Linear acceleration (a) = angular acceleration (a^, ) x radius

a_{1 =} a^,  x r₁

₌  0.03924 x 49 x 10^-2   = 0.0192276 m/sec² ( upward )

a_{2 =} a^,  x r₂

₌  0.03924 x 20 x 10^-2   = 0.007848 m/sec² ( downward , hence take with - ve sign )

Thank you .