Answer: 24.49 m/s
Explanation:
Given
Mass of hoop, m = 2 kg
Radius of hoop, r = 0.5 m
Speed of rolling, v = 15 m/s
Work done, = 750 J
If we assume that the hoop is not skidding, then it's Kinetic Energy of rotation is 1/2mv²
also the Kinetic Energy of translation is 1/2mv², thus the total Kinetic Energy is
KE = KE(r) + KE(t)
KE = 1/2mv² + 1/2mv²
KE = mv²
Recall, the change in kinetic energy = work done = 750 J
mv² - mv•² = 750
m (v² - v•²) = 750
v² - v•² = 750/m
v² - 15² = 750/2
v² = 375 + 225
v² = 600
v = 24.49 m/s
Therefore, the new speed is 24.49 m/s
The new speed of the center of mass of the hoop is 27.38 m/s.
Given data:
The mass of hoop is, m = 2.0 kg.
The radius of hoop is, r = 0.50 m.
The speed of center of mass is, u = 15 m/s.
The work done by the external force is, W = 750 J.
The given problem is based on the total energy during the linear motion. Generally when a body is rolling in a linear path, then the total kinetic energy is equal to sum of translational kinetic energy and rotational kinetic energy.
First apply the work -energy concept as,
[tex]W =\Delta KE\\\\W =\dfrac{1}{2}m(v^{2}-u^{2})+\dfrac{1}{2}(I \times \omega^{2})[/tex] ..............................................................(1)
Here,
v is the new speed of center of mass.
[tex]\omega[/tex] is the angular speed and I is the moment of inertia.
Solving as,
[tex]W =\dfrac{1}{2}m(v^{2}-u^{2})+\dfrac{1}{2}(mr^{2} \times (u / r )^{2})\\\\W =\dfrac{1}{2}m(v^{2}-u^{2})+\dfrac{1}{2}(mu^{2} )\\\\W =\dfrac{1}{2}mv^{2}-\dfrac{1}{2}mu^{2}+\dfrac{1}{2}mu^{2}\\\\W =\dfrac{1}{2}mv^{2}[/tex]
Substituting the values as,
[tex]750 =\dfrac{1}{2} \times 2 \times v^{2}\\\\v = \sqrt{750}\\\\v = 27.38 \;\rm m/s[/tex]
Thus, we can conclude that the new speed of the center of mass of the hoop is 27.38 m/s.
Learn more about the work energy theorem here:
https://brainly.com/question/24509770
