1. One radian is about:
Verification Matrix:
Since a full circle is $360^\circ$ or $2\pi\text{ radians}$, we calculate $1\text{ radian} = \frac{360^\circ}{2\pi} \approx \frac{360^\circ}{6.283} \approx 57.3^\circ$.
2. If a wheel turns with constant angular speed, then:
Verification Matrix:
By definition, constant angular speed ($\omega = \frac{\Delta\theta}{\Delta t}$) means the object describes equal angular displacements in equal intervals of time. Points on the rim change direction continuously, meaning their linear velocity and acceleration vectors are not constant.
3. The rotational inertia (moment of inertia) of a wheel about its axle does not depend upon its:
Verification Matrix:
Moment of inertia ($I = \sum m_i r_i^2$) is strictly a geometric structural property determined by the total mass, dimensions (diameter), and how that mass is distributed relative to the axis of rotation. It is entirely independent of kinematic variables like its rotational speed ($\omega$).
4. A force with a given magnitude is to be applied to a wheel. The torque can be maximized by:
Verification Matrix:
Torque is given by $\tau = r F \sin\theta$. To maximize $\tau$ for a fixed force $F$, you must maximize the moment arm $r$ (by moving out to the rim) and maximize $\sin\theta$ (by applying it perpendicularly, i.e., tangent to the rim where $\theta = 90^\circ, \sin90^\circ = 1$).
5. For an object rotating about a fixed axis, where $I$ is its rotational inertia and $\alpha$ is its angular acceleration, $\tau = I\alpha$:
Verification Matrix:
$\tau = I\alpha$ is the rotational analog of Newton's Second Law ($F = ma$). It is derived directly by applying $F = ma$ to each individual mass element of a rigid body undergoing angular acceleration.
6. The angular momentum vector of Earth about its rotation axis, due to its daily rotation, is directed:
Verification Matrix:
Applying the **Right-Hand Rule**: Curl the fingers of your right hand in the direction of Earth's rotation (West to East). Your extended thumb points along the rotational axis toward the geographic North Pole.
7. A stone of $2\text{ kg}$ is tied to a $0.50\text{ m}$ long string and swung around a circle at an angular velocity of $12\text{ rad/s}$. The net torque on the stone about the center of the circle is:
Verification Matrix:
Since the stone is moving at a *constant* angular velocity ($\omega = 12\text{ rad/s}$), its angular acceleration is zero ($\alpha = 0$). Because torque depends on angular acceleration ($\tau = I\alpha$), the net torque acting on the system is exactly $0\text{ N}\cdot\text{m}$.
8. A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms:
Verification Matrix:
Because the system is frictionless, no external net torque acts on it ($\tau_{\text{ext}} = 0$), meaning total angular momentum is strictly conserved ($L = I\omega = \text{constant}$). When he extends his arms, his mass distribution moves outward, causing rotational inertia ($I$) to *increase* and angular velocity ($\omega$) to *decrease*, while $L$ stays unchanged.
9. A space station revolves around the earth as a satellite, $100\text{ km}$ above the Earth's surface. What is the net force on an astronaut at rest inside the space station?
Verification Matrix:
The sensation of weightlessness occurs because the astronaut and the space station are in a continuous state of free fall together. However, a real gravitational pull still acts on her! Since $100\text{ km}$ is a small distance compared to the Earth's radius ($\approx 6400\text{ km}$), gravity at this altitude is only slightly weaker, making the actual net gravitational force *a little less than her weight on earth*.
10. If the external torque acting on a body is zero, then its:
Verification Matrix:
From the rotational framework of impulse and momentum: $\tau_{\text{ext}} = \frac{\Delta L}{\Delta t}$. If $\tau_{\text{ext}} = 0$, then $\Delta L = 0$, proving that total angular momentum remains strictly conserved over time.
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