A student sits on a rotating stool holding two 5 kg objects. When his arms are extended horizontally,

A student sits on a rotating stool holding two 5 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.73 rad/sec. The moment of inertia of the student plus the stool is 2 kg m^2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.31 m from the rotation axis.
(a) Calculate the final angular speed of the student. Answer in units of rad/s.
(b) Find the kinetic energy of the student before and after the objects are pulled in.

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  1. A) ω2 = 1.394 rad/s

    B) KE_i = 1.7085 J and KE_f = 3.2196J

    Explanation:

    We are given;

    Mass of objects: m1 = m2 = 2kg

    Radius: r1 = 0.9m ; r2 = 0.28m

    Angular velocity: ω1 = 0.74 rad/s

    Moment of inertia of stool and object: I = 3kg.m²

    A) Using the law of conservation of angular momentum:

    I1•ω1 = I2•ω2

    The values of the moment of inertia according to the parallel axis theorem will be:

    I1 = I + 2mr1²

    So, I1 = 3 + 2(2 x 0.9²)

    I1 = 6.24 kg.m²

    Similarly,

    I2 = I + 2mr2²

    I2 = 3 + 2(2)(0.28²)

    I2 = 3.3136 kg.m²

    Since I1•ω1 = I2•ω2

    Thus, ω2 = I1•ω1/I2

    ω2 = 6.24 x 0.74/3.3136

    ω2 = 1.394 rad/s

    B) The initial kinetic energy is:

    KE_i = (1/2)•I1•ω1²

    KE_i = (1/2) x 6.24 x 0.74²

    KE_i = 1.7085 J

    The final kinetic energy is;

    KE_f = (1/2)•I2•ω2²

    KE_f = (1/2) x 3.3136 x 1.394²

    KE_f = 3.2196J

  2. Explanation:

    Given that,

    Two object of Mass = 5kg

    M1 = M2 = 5kg

    Radius of rotation r = 1.1m

    Angular speed wi = 0.73rad/s

    Moment of inertia I = 2kgm²

    The student pull the mass to another position of r' = 0.31m

    A. Final angular momentum?

    Using the conservation of angular momentum

    The momentum of inertia of the objects the students holding I = mr²

    L(initial) = L(final)

    I•wi = I•wf

    (I+mr²+mr²)•wi = (I+mr'²+mr'²)•wf

    (2+5•1.1²+5•1.1²)•0.73=(2+5•0.31²+5•0.31²)wf

    10.293 = 2.961wf

    Then, wf= 10.293/2.961

    wf = 3.476 rad/s

    wf ≈ 3.48rad/s

    b) kinetic energy before and after

    Kinetic energy is give as

    Kinetic energy = ½•I•w²

    Initial kinetic energy is

    K.E(initial) = ½(I+mr²+mr²)wi²

    K.E(initial) = ½(2+5•1.1²+5•1.1²)•0.73²

    K.E(initial) = 3.76 J

    Final kinetic energy is

    K.E(final) = ½(I+mr'²+mr'²)wf²

    K.E(final)=½(2+5•0.31²+5•0.31²)3.48²

    K.E(final) = 17.93J

  3. a) the final angular speed is 0.738 rad/s

    b) the change in kinetic energy = 0.3 J

    Explanation:

    the two 1 kg objects have a total mass of 2 x 1 = 2 kg

    radius of rotation of the objects = 0.9 m

    moment of inertial of the student and the chair = 6 kg-m^2

    initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s

    final angular speed of rotation of the sitting student and object system ω2 = ?

    moment of inertia of the rotating object is

    [tex]I = mr^{2}[/tex] = 2 x [tex]0.9^{2}[/tex] = 1.62 kg-m^2

    total moment of inertia of sitting student and object system will be  

    ==> 6 + 1.62 = 7.62 kg-m^2

    The initial angular momentum of the sitting student and object system will be calculated from

    ==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2

    if the radius of rotation of the object is reduced to 0.39 m,

    new moment of inertia of the rotating object will be

    [tex]I = mr^{2}[/tex]  = 2 x [tex]0.39^{2}[/tex] = 0.304 kg-m^2

    new total moment of inertia of the sitting student and object system will be

    ==> 6 + 0.304 = 6.304 kg-m^2

    The final momentum of the sitting student and object system will be calculated from

    ==> Iω2 = 6.304 x ω2 = 6.304ω2

    According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system. Therefore,

    4.65 = 6.304ω2

    ω2 = 4.65/6.30 = 0.738 rad/s

    b) Rotational kinetic energy of the system = [tex]\frac{1}{2} Iw^{2}[/tex]

    for the initial conditions, kinetic energy is

    ==>  [tex]\frac{1}{2} Iw1^{2}[/tex] =  [tex]\frac{1}{2}* 7.62*0.61^{2}[/tex] = 1.417 J

    for the final conditions, kinetic energy is

    ==>  [tex]\frac{1}{2} Iw1^{2}[/tex] =  [tex]\frac{1}{2}*6.304*0.738^{2}[/tex] = 1.717 J

    change in kinetic energy = final KE - initial KE

    ==> 1.717 - 1.417 = 0.3 J

  4. [tex]\omega _f=1.185\ rad/s[/tex]

    Explanation:

    Given

    mass of objects [tex]m=5\ kg[/tex]

    Initially mass is at [tex]r=0.9\ m[/tex]

    Initial angular speed [tex]\omega_i=0.66\ rad/s[/tex]

    Moment of inertia of student and  stool is [tex]I_s=8\ kg-m^2[/tex]

    Finally masses are at a distance of [tex]r_f=0.31\ m[/tex] from axis

    [tex]I_i=I_p+I_m[/tex]

    [tex]I_i=8+2\times 5\times (0.9)^2[/tex]

    [tex]I_i=16.1\ kg-m^2[/tex]

    Final moment of inertia of the system

    [tex]I_f=I_s+I_m[/tex]

    [tex]I_f=8+2\times 5\times (0.31)^2[/tex]

    [tex]I_f=8+0.961=8.961\ kg-m^2[/tex]

    As there is no external torque therefore moment of inertia is conserved

    [tex]I_i\omega _i=I_f\omega _f[/tex]

    [tex]\omega _f=\frac{16.1}{8.96}\times 0.66[/tex]

    [tex]\omega _f=1.796\times 0.66[/tex]

    [tex]\omega _f=1.185\ rad/s[/tex]

  5. Only part of your force does work because your force has both a horizontal part and a vertical part. When you pull this way, only part in the same direction as the motion of the suitcase does work. ... The force you exert and the distance that the object moves.

  6. Answers:

    a) [tex]1.05 rad/s[/tex]

    b) [tex]1.38 J[/tex]

    Explanation:

    a) Final angular velocity :

    Before solving this part, we have to stay clear that the angular momentum [tex]L[/tex] is conserved, since we are dealing with circular motion, then:

    [tex]L_{o}=L_{f}[/tex]

    Hence:

    [tex]I_{i} \omega_{i}=I_{f} \omega_{f}[/tex](1)

    Where:

    [tex]I_{i}[/tex] is the initial moment of inertia of the system

    [tex]\omega_{i}=0.73 rad/s[/tex] is the initial angular velocity

    [tex]I_{f}[/tex] is the final moment of inertia of the system

    [tex]\omega_{f}[/tex] is the final angular velocity

    But first, we have to find [tex]I_{i}[/tex] and [tex]I_{f}[/tex]:

    [tex]I_{i}=I_{s}+2mr_{i}^{2}[/tex](2)

    [tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex](3)

    Where:

    [tex]I_{s}=8 kgm^{2}[/tex] is the student's moment of inertia

    [tex]m=2 kg[/tex] is the mass of each object

    [tex]r_{i}=1 m[/tex] is the initial radius

    [tex]r_{f}=0.28 m[/tex] is the final radius

    Then:

    [tex]I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}[/tex](4)

    [tex]I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}[/tex](5)

    Substituting the results of (4) and (5) in (1):

    [tex](12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}[/tex](6)

    Finding [tex]\omega_{f}[/tex]:

    [tex]\omega_{f}=1.05 rad/s[/tex]  (7) This is the final angular speed

    b) Change in kinetic energy:

    The rotational kinetic energy is defined as:

    [tex]K=\frac{1}{2}I \omega^{2}[/tex](8)

    And the change in kinetic energy is:

    [tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{i} \omega_{i}^{2}[/tex](9)

    Since we already calculated these values, we can solve (9):

    [tex]\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}[/tex](10)

    Finally:

    [tex]\Delta K=1.38 J[/tex]This is the change in kinetic energy

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