Comments (10) on “A wheel of radius 0.5 m rotates with a constant angular speed about an axis perpendicular to its center.”

  3. [tex]\omega=2.85*10^{13}\frac{rad}{s}[/tex]

    Explanation:

    The translational kinetic energy depends on the mass and speed of the body, as follows:

    [tex]K_T=\frac{mv^2}{2}\\K_T=\frac{5.30*10^{-26}kg(1.49*10^3\frac{m}{s})^2}{2}\\\\K_T=1.18*10^{-19}J[/tex]

    While rotational kinetic energy depends on the moment of inertia and the angular velocity of the body, as follows:

    [tex]K_R=\frac{I\omega^2}{2}(1)[/tex]. We know that:

    [tex]K_R=\frac{2}{3}K_T(2)[/tex]

    Replacing (1) in (2):

    [tex]\frac{I\omega^2}{2}=\frac{2}{3}K_T\\\\\omega=\sqrt{\frac{4}{3}\frac{K_T}{I}}\\\omega=\sqrt{\frac{4}{3}\frac{1.18*10^{-19}J}{1.94*10^{-46}kg\cdot m^2}}\\\omega=2.85*10^{13}\frac{rad}{s}[/tex]

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  4. The tangential acceleration is 0 m/s².

    Explanation:Given:

    Radius of the wheel = 0.5 m

    The point of observation for calculating tangential acceleration = 0.2 m from center.

    Tangential speed at the point of observation = 2 m/s

    The angular speed of the wheel is a constant.

    In order to determine the tangential acceleration, we make use of the following formula:

    Tangential acceleration at a point = Angular acceleration × Distance of the point from center

    Or, a_t=\alpha \times r

    Now, angular acceleration is defined as the rate of change of angular speed.

    Here, the angular speed of the wheel is a constant. So, the change of angular speed is 0. Therefore, the angular acceleration is also 0 rad/s².

    Now, from the above formula, as angular acceleration is 0, the magnitude of tangential acceleration at a point that is 0.2 m from the center of the wheel is also 0 m/s².

    Hope it helped

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  7. Remark

    You can use  a method called unit analysis to solve this problem. You will use it many times in your education later on.

    Solve

    [tex]\dfrac{28 mile}{hour} *\dfrac{5280 feet}{mile}*\dfrac{1 hour}{60 min}*\dfrac{1 min}{60 sec}[/tex]

    Now sort out the numbers  in the numerator and those in the denominator and multiply them together.

    Numerator 28 * 5280 * 1 * 1 = 147840

    Denominator: 60 * 60 = 3600

    Now divide the numerator by denominator

    Answer = 147840 / 3600 = 41.1 Feet / second

    28 miles per hour = 41.1 feet / second

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  8. Option B is correct.

    Explanation:

    Earth rotates on its axis at an average speed of 1700 km/hr. That seems like a really high speed, however, we orbit the sun at around 107.000 km/hr.

    And going even further, our entire solar system orbits the Galactic Center of the Milky Way at an average of 828.000 km/h.

    Our entire galaxy also moves, although its hard to determine its exact speed. Scientists believe that the Milky way moves at 2.2 million km/h relative to the CBR (Cosmic Background Radiation) that its present everywhere across the universe.

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  9. a) 120 s

    b) v = 0.052R [m/s]

    Explanation:

    a)

    The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

    The graph of the problem is missing, find it in attachment.

    To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

    The first point we take is t = 0, when the position of the book is x = 0.

    Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

    Therefore, the period is

    T = 120 s - 0 s = 120 s

    b)

    The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

    The perimeter of the wheel is:

    [tex]L=2\pi R[/tex]

    where R is the radius of the wheel.

    The period of revolution is:

    [tex]T=120 s[/tex]

    Therefore, the tangential speed of the book is:

    [tex]v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R[/tex]

    [tex]A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass m[/tex]

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