Question

please answer fully with work shown! Thank you:)

2. Refer to the diagram shown here (a side view, not to scale Two blocks with masses m, and m,. respectively, are connected via a massless thread over a pulley wheel. The wheel is a solid, uniform disk with mass m,, outer radius R, and a frictionless axle Initially you hold m, (and therefore the entire system) at rest in the position shown; m, is at rest on level ground. Then you release m2 The data: m,-7.16 kg m-345 kg m,-2.63 kg R-5.10 cm H 27.4 cm g 9.80 m/s2 By what percentage does the tension force acting on m, change when you release m,2 How fast, measured in rpm, is the wheel spinning when m, a. b. reaches the surface? You may ignore air drag. (ground)

Answer 1

When the mass m is kept hold the mass m, will be at rest Therefore, -(7.16 kg)(9.81m/s') -70.239 N When the mass m is released

T, 43 m2g 71 m1g Now, apply the Newton's laws. migi ma

The torque acting on the pulley 1S, myR'a 2R Substitute, the equations (1) and (2) in equation (3) 7-不一 m,Ra (m-m2)g 43 1m m1 + m2 +-- Substitute, the values in the above equation.

(7.16 kg -3.45 kg) (9.81 m/s 2.63 kg 7.16 kg +3.45 kg + -3.052 m/s The tension is, (7.16 kg) (9.81 m/s1-3.052 m/s -48.38 N ow. - 70.239 N-48.38 N -21.85 N The percentage of the tension when m, is released is

%tension (T1), %tension 100% 1 2.85 N 70.239 N 31.10% -- x 10090 The time taken by the block to reach the ground is, DER 2(0.274 m) 13.052 m/s -0.423 s The angular acceleration is, 3.052 m/s 0.0510 m 59.843 rad/s2

The angular velocity is, -(59.843 rad/s') (0.423 s) -14.122 rad/s -242.27 rpm 1123 3 60