You have two solid steel spheres. Sphere 2 has twice the radius of sphere 1. By what factor does the moment of inertia I2 of sphere 2 exceed the moment of inertia I1 of sphere 1?

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Robert Pina · Accepted Answer

Firstly, I∝mr2    m=43πr3ρsteel    As sphere 2 has twice the radius, same destiny&amp;lt; its mass is greater by a factor of 23=8    The extra mass is also distributed farther from the center (r2=2r1), thus,    I∝mr2⇒I2∝(8m1)(2r1)2=32I1

Eliza Beth13 · Accepted Answer

To solve the problem, we need to find the factor by which the moment of inertia of sphere 2 exceeds the moment of inertia of sphere 1. Let&#039;s denote the radius of sphere 1 as r1 and the radius of sphere 2 as r2.Given that sphere 2 has twice the radius of sphere 1, we have r2=2r1.The moment of inertia of a solid sphere is given by the formula:I=25mr2,where m is the mass of the sphere and r is its radius.Since both spheres are made of solid steel, the masses of the spheres will be proportional to their volumes. The volume of a sphere is given by:V=43πr3.So, the mass m of a sphere will be proportional to r3.Let&#039;s assume the mass of sphere 1 is m1, and the mass of sphere 2 is m2. We can write the following equation:m2m1=(r2r1)3.Substituting r2=2r1 into the equation, we have:m2m1=(2r1r1)3=23=8.So, the mass of sphere 2 is eight times the mass of sphere 1.Now, let&#039;s find the ratio of the moments of inertia:I2I1=25m2r2225m1r12=m2r22m1r12.Substituting m2=8m1 and r2=2r1 into the equation, we have:I2I1=(8m1)(2r1)2m1r12=32m1r12m1r12=32.Therefore, the moment of inertia I2 of sphere 2 exceeds the moment of inertia I1 of sphere 1 by a factor of 32.

Don Sumner · Accepted Answer

We are given two solid steel spheres, where Sphere 2 has twice the radius of Sphere 1. Let&#039;s denote the radius of Sphere 1 as r1 and the radius of Sphere 2 as r2.The moment of inertia of a solid sphere is given by the formula:I=25mr2,where m is the mass of the sphere and r is its radius.Since both spheres are made of solid steel, their densities are the same. Hence, their masses will be proportional to their volumes. The volume of a sphere is given by:V=43πr3.Thus, the mass m of a sphere will be proportional to r3.Let&#039;s assume the mass of Sphere 1 is m1, and the mass of Sphere 2 is m2. We can write the following equation:m2m1=(r2r1)3.Since Sphere 2 has twice the radius of Sphere 1, we have r2=2r1. Substituting this into the equation, we get:m2m1=(2r1r1)3=23=8.Thus, the mass of Sphere 2 is eight times the mass of Sphere 1.Now, let&#039;s find the ratio of the moments of inertia:I2I1=25m2r2225m1r12=m2r22m1r12.Substituting m2=8m1 and r2=2r1 into the equation, we have:I2I1=(8m1)(2r1)2m1r12=32m1r12m1r12=32.Therefore, the moment of inertia I2 of Sphere 2 exceeds the moment of inertia I1 of Sphere 1 by a factor of 32.

madeleinejames20 · Accepted Answer

Answer: 4Explanation:I=25mr2where I is the moment of inertia, m is the mass of the sphere, and r is the radius of the sphere.Let&#039;s assume the mass of both spheres is the same. Therefore, we can compare the moments of inertia by using the ratio:I2I1=25m(2r1)225mr12Simplifying the equation, we have:I2I1=25(2r1)225r12I2I1=254r1225r12I2I1=4r12r12I2I1=4Therefore, the moment of inertia I2 of sphere 2 exceeds the moment of inertia I1 of sphere 1 by a factor of 4.

You have two solid steel spheres. Sphere 2 has twice

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