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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

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cjrodriguez89

Answer:

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

Explanation:

a)

  • Assuming no external torques act on the skater, total angular momentum must be conserved:

        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)

  • As the final angular momentum must be the same, and we know the value of the final angular velocity, we can replace by the values in L2, and solve for I2, as follows:

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

c)

  • If an external torque is present, we can write the following equation, that relates the external torque with the rotational inertia and the angular acceleration, as follows:

        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

Answer:

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

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