5. . An point mass (2 Kg) is moving with position vector r(t) = 3t²i? - 2tj? m with respect to origin O (0,0) , and time : t . Find the angular momentum L? of the object about point O , at time t=2 s. (a) 40 k? Kg m²/s (b) 80 k? Kg m/s (c) 80 k? Kg m²/s (d) none of them solutions (d) . Quiz-3 , question no 31 (radius vector is changed) , also discussed in during lecture and recitation. L? = r? × p? , r? = (12i? - 4j? ) m at t=2 s , v? = (6ti? - 2j?), p? = mv? = (24i? - 4j?) (Kg m/s) at 2s, Thus L? = r? × p? = [(12)(-4) - (-4)(24)]k? = 48k? (Kg m²/s) 6. for a conical pendulum rotating with constant speed in circular motion : which one below are true ? (angular momentum is calculated about the point A) : figure 3 (a) angular momentum is conserved about point A (b) magnitude of angular momentum and direction both changes with time (c) magnitude is constant but direction changes (d) none of the options are correct Figure 3 solutions (c) . Quiz-3 , question no 32 (options are rearranged) , also discussed in during lecture and recitation , Torque due to tension T about point 'A' is zero , however Torque due to mg about 'A' is not zero. Thus we have net torque , so angular momentum vector can not be conserved, however its magnitude is about point 'A' .
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Given position vector r = (7i + 32j - 2tj), we can find the velocity vector v by differentiating the position vector with respect to time. v = dr/dt = -2j So, the velocity vector at t = 2 seconds is v = -2j. Show more…
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A 500 -g uniform sphere of $7.0$ -cm radius spins frictionlessly at 30 rev/s on an axis through its center. Find its ( $a$ ) $\mathrm{KE}_{r}$, (b) angular momentum, and ( $c$ ) radius of gyration. We need the moment of inertia of a uniform sphere about an axis through its center. From $\underline{\text { Fig. } 10-1}$, $$I=\frac{2}{5} M r^{2}=(0.40)(0.50 \mathrm{~kg})(0.070 \mathrm{~m})^{2}=0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ (a) Knowing that $\omega=30 \mathrm{rev} / \mathrm{s}=188 \mathrm{rad} / \mathrm{s}$, we have $$\mathrm{KE}_{r}=\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})^{2}=0.017 \mathrm{~kJ}$$ Notice that $\omega$ must be in $\mathrm{rad} / \mathrm{s}$. (b) Its angular momentum is $$L=I \omega=\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})=0.18 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$ (c) For any object, $I=M k^{2}$, where $k$ is the radius of gyration. Therefore, $$k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}}{0.50 \mathrm{~kg}}}=0.044 \mathrm{~m}=4.4 \mathrm{~cm}$$ Notice that this is a reasonable value in view of the fact that the radius of the sphere is $7.0 \mathrm{~cm}$.
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1. (10 points) The angular velocity of a rotating disk is given by ω(t) = 1.6t³ - 4.2t² (rad/s). What is the average angular acceleration of the disk between t = 1 s and t = 3 s? Answer: 4 rad/s² 2. (10 points) The angular position of a point on a rotating wheel is given by θ(t) = t⁴ - 4t² + 6, where θ is in radians and t is in seconds. What is the point's angular acceleration at t = 2 s? Answer: 40 rad/s² 3. (10 points) Time dependent angular velocity of a particle is given by ω(t) = 5t² + 6t + 1 (rad/s). What is the average angular acceleration in the time interval t=0 to t=2s? Answer: αₐᵫₒ = 16 rad/s² 4. (10 points) Two point particles are fastened to a rod of negligible mass as shown in the figure. If the rotational inertia about an axis that is perpendicular to the rod and 20 cm away from the first particle is I = 0.09 kg.m², what is the distance x₂ from second particle to the rotation axis? Answer: x₂ = 0.05m 5. (10 points) A rotor has a moment of inertia of 4.20×10⁻² kg.m². How much energy is required to bring it from rest to 9300 rpm (rev/min) speed?
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0 (a) The disc experiences gravity, the force of reaction of the horizontal surface, and the force $\vec{R}$ of reaction of the wall at the moment of the impact against it. The first two forces counter-balance each other, leaving only the force $\vec{R}$. It's moment relative to any point of the line along which the vector $\vec{R}$ acts or along normal to the wall is equal to zero and therefore the angular momentum of the disc relative to any of these points does not change in the given process. (b) During the course of collision with wall the position of disc is same and is equal to $\overrightarrow{r_{o o^{\prime}}}$ Obviously the increment in linear momentum of the ball $\Delta \vec{p}=2 m v \cos \alpha \hat{n}$ Here, $\Delta \vec{M}=\vec{r}_{o o^{\prime}} \times \Delta \vec{p}=2 m v l \cos \alpha \hat{n}$ and directed normally emerging from the plane of figure Thus $|\Delta \vec{M}|=2 m v l \cos \alpha$
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