Question

Five hoops are each pivoted at a point on the rim and allowed to swing as physical pendulums. The masses and radii are hoop 1: M = 150 g and R = 50 cm hoop 2: M = 200 g and R = 40 cm hoop 3: M = 250 g and R = 30 cm hoop 4: M = 300 g and R = 20 cm hoop 5: M = 350 g and R = 10 cm Order the hoops according to the periods of their motions, smallest to largest.

Answer 1

The order of the hoops according to the periods of their motions, from smallest to largest, is Hoop1, Hoop2, Hoop3, Hoop5, Hoop4. Thus, the correct answer is option C. 1, 2, 3, 5, 4,

Step-by-step explanation:

The period of a physical pendulum is given by the formula
`\(T = 2π\sqrt{(I)/(mgd)}\)`, where I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot to the center of mass. For a hoop, I is
`\(MR^2\)`. The period is inversely proportional to the square root of I, so larger I values lead to longer periods.

Calculating the periods for each hoop, we find that Hoop 1 has the smallest period and Hoop 5 has the largest period:

`\[T_1 \propto (R_1)/(√(g))\]`

`\[T_2 \propto (R_2)/(√(g))\]`

`\[T_3 \propto (R_3)/(√(g))\]`

`\[T_4 \propto (R_4)/(√(g))\]`

`\[T_5 \propto (R_5)/(√(g))\]`

Comparing the ratios, we see that
`\(T_1 < T_2 < T_3 < T_5 < T_4\).`Therefore, the correct order from smallest to largest period is 1, 2, 3, 5, 4, leading to the final answer C. 1, 2, 3, 5, 4.

Full Question:

Five hoops are each pivoted at a point on the rim and allowed to swing as physical pendulums. The masses and radii are hoop 1: M = 150g and R = 50cm hoop 2: M = 200g and R = 40cm hoop 3: M = 250g and R = 30cm hoop 4: M = 300g and R = 20cm hoop 5: M = 350g and R = 10cm Order the hoops according to the periods of their motions, smallest to largest.A. 1, 2, 3, 4, 5 B. 5, 4, 3, 2, 1 C. 1, 2, 3, 5, 4 D. 1, 2, 5, 4, 3 E. 5, 4, 1, 2, 3