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levacccp [35]
3 years ago
10

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

re (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of → r and → F ?
Physics
2 answers:
andrey2020 [161]3 years ago
8 0

Explanation:

It is given that,

Force, F=(-8\ N)i+(6\ N)j

Position vector, r=(3i+4j)\ m

(a) The torque on the particle about the origin is given by :

\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m

(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

snow_lady [41]3 years ago
7 0

Answer:

Explanation:

\overrightarrow{F}=-8\widehat{i}+6\widehat{j}

\overrightarrow{r}=3\widehat{i}+4\widehat{j}

(a) the formula for the torque is given by

\overrightarrow{\tau }=\overrightarrow{r} \times \overrightarrow{F}

\overrightarrow{\tau }=\left (3\widehat{i}+4\widehat{j}  \right )\times \left (-8\widehat{i}+6\widehat{j}  \right )

\overrightarrow{\tau }=50\widehat{k}

(b) Let the angle between r and F is θ.

Using the formula

Cos\theta =\frac{\overrightarrow{F}.\overrightarrow{r}}{\mid \overrightarrow{r}\mid .\mid \overrightarrow{F}\mid }

\mid \overrightarrow{r}\mid = \sqrt{3^{2}+4^{2}}=5

\mid \overrightarrow{F}\mid = \sqrt{8^{2}+6^{2}}=10

\overrightarrow{F}.\overrightarrow{r} = \left ( -8\widehat{i}+6\widehat{j} \right ).\left ( 3\widehat{i}+4\widehat{j} \right )=-24+24=0

So, Cos θ = 0

θ = 90°

Thus, the angle between f and r is 90°.

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3 years ago
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/
german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = \tau \theta = I\alpha * \theta

I is the rotational inertia = 16kgm²

\theta = angular\ displacement

\theta = 2 rev = 12.56 rad

\alpha \ is \ the\ angular\ acceleration

To get the angular acceleration, we will use the formula;

\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}

\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

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8 0
3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

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