Calculate the angular momentum, in kg · m2/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.310 kg · m2. kg · m2/s (b) he reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. find the value of his moment of inertia, in kg · m2, if his angular velocity drops to 2.45 rev/s. kg · m2 (c) suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. what average torque, in n m, was exerted if this takes 18.0 seconds? (indicate the direction with the sign of your answer. assume that the skater's rotation is in the positive direction.) n · m

a)11.7 kg. m2/s b) 0.76 Kg. m2 c) -0.33 N.m

Explanation:

a)

        L1 = L2

        As the angular momentum can be calculated as the

        product of the moment of inertia times the angular velocity,

we can write:

       I1*ω1 = I2*ω2  

       The initial angular momentum can be written as follows:

       I1*ω1= 0.31 kg.m2 * 6.0 rev/sec  

       As we need to express the angular momentum in kg.m2/s, we need to convert the angular velocity units, from rev/sec to rad/sec, as follows:

       ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec

       I1*ω1= 0.31 kg.m2 * 12 π rad/sec = 11.7 kg. m2/s

b)

        I2 = I1*( ω1 / ω2) = 0.31 kg. m2 . 6.0/2.45 = 0.76 kg.m2

c)

        Τ= I *γ (1)

        Where γ, is the angular acceleration.

        By definition, γ is the rate of change of the angular velocity,

        so if we have the values of  the initial and final angular

        velocities, and the time passed, we can express γ as

follows:

       γ= (ω2 – ω1) / t

       In order to express γ in rad/sec2, we need to convert the

angular  velocities (given in rev/sec), to rad/sec, as follows:

       ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec

       ω2= 3.0 rev/sec (2π rad/ rev) = 6 π rad/sec

       Solving for γ:

       γ = -6 π / 18. 0 rad/sec2 = -1.05 rad/sec2

       Replacing in (1), we have:

       τ= 0.31 kg. m2.*(-1.05 rad/sec2) = -0.33 N.m

a) The angular momentum is 11.69 kg m²/s

b) The moment of inertia is 0.76 kg m²

c) The torque is -0.3255 N

Explanation:

please look at the solution in the attached Word file

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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