The disk with radius of R=2m and mass of m=5kg rotates like a merry-go-round around its central axis with constant speed of 33 rad/s. The bird lands on the disk at 1 m from disk's center then bird and disk rotate together with angular speed of 32rad/s. Determine the mass of bird. [ Treat bird as a particle. I_(diskcom)=1/2mR^2 Select one: a. 1.3 kg b. 0.5 kg c. 0.63 kg d. 0.31 kg e. 1.0 kg

To solve this problem, we can use the conservation of angular momentum.<br /><br />Given information:<br />- Radius of the disk, R = 2 m<br />- Mass of the disk, m = 5 kg<br />- Constant speed of the merry-go-round, ω1 = 33 rad/s<br />- Distance of the bird from the center of the disk, r = 1 m<br />- Angular speed of the bird and disk together, ω2 = 32 rad/s<br /><br />Step 1: Calculate the moment of inertia of the disk.<br />$I_{disk} = \frac{1}{2} mR^2$<br />$I_{disk} = \frac{1}{2} \times 5 \times (2)^2$<br />$I_{disk} = 10 \, \text{kg} \cdot \text{m}^2$<br /><br />Step 2: Calculate the total angular momentum before the bird lands on the disk.<br />$L_{initial} = I_{disk} \times ω1$<br />$L_{initial} = 10 \times 33$<br />$L_{initial} = 330 \, \text{kg} \cdot \text{m}^2/\text{s}$<br /><br />Step 3: Calculate the total angular momentum after the bird lands on the disk.<br />$L_{final} = (I_{disk} + I_{bird}) \times ω2$<br />$I_{bird} = m_{bird} \times r^2$<br />$L_{final} = (10 + m_{bird} \times 1^2) \times 32$<br />$L_{final} = (10 + m_{bird}) \times 32$<br /><br />Step 4: Apply the conservation of angular momentum.<br />$L_{initial} = L_{final}$<br />$330 = (10 + m_{bird}) \times 32$<br />$m_{bird} = \frac{330}{32} - 10$<br />$m_{bird} = 0.3125 \, \text{kg}$<br /><br />Therefore, the mass of the bird is approximately 0.31 kg.<br /><br />The correct answer is:<br />d. 0.31 kg