It is necessary to determine the normal acceleration of a point at a given point in time, provided that the acceleration of the point is 1.5 m/s² and the angle between the acceleration and velocity vectors is 65°. Round your answer to two decimal places.
Answer:
It is known that the vector product of the velocity and acceleration of a point is equal to the projection of the acceleration onto the direction of the radius of curvature:
$$\vec{v} \times \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(v\vec{e}_t) = \frac{ dv}{dt}\vec{e}_t + v\frac{d\vec{e}_t}{dt} = \frac{dv}{dt}\vec{e}_t + v^2\vec{e }_n,$$
where $\vec{e}_t$ and $\vec{e}_n$ are the unit vectors tangent and normal to the curve, respectively.
The normal acceleration of a point is defined as the modulus of the vector of the radius of curvature multiplied by the square of the velocity:
$$a_n = |\vec{a}_n| = \frac{|\vec{v} \times \vec{a}|}{v^2}.$$
From the conditions of the problem, the acceleration of the point and the angle between the acceleration and velocity vectors are known:
$$a = 1.5\ m/s^2,$$
$$\theta = 65^{\circ}.$$
Therefore, the acceleration of a point can be decomposed into tangent and normal:
$$\vec{a} = \vec{a}_t + \vec{a}_n,$$
where $\vec{a}_t$ and $\vec{a}_n$ are the tangential and normal accelerations, respectively.
The angle between vectors $\vec{a}$ and $\vec{v}$ is equal to $90^{\circ} - \theta$, therefore, the projection of acceleration onto the direction of the radius of curvature:
$$|\vec{a}_n| = |\vec{a}|\sin(90^{\circ} - \theta) = a\sin\theta = 1.5\ м/с^2 \cdot \sin 65^{\circ} \approx ,36\ м/с^2.$$
Thus, the normal acceleration of a point at a given time is approximately 1.36 m/s².
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We present to your attention the solution to problem 7.8.4 from the collection of Kepe O.?. in physics.
First, let's calculate the projection of acceleration onto the direction of the radius of curvature using the formula:
$$|\vec{a}_n| = |\vec{a}|\sin(90^{\circ} - \theta) = a\sin\theta = 1.5\ м/с^2 \cdot \sin 65^{\circ} \approx ,36\ м/с^2.$$
Then we calculate the normal acceleration of a point at a given time using the formula:
$$a_n = |\vec{a}_n| / v^2,$$
where $v$ is the speed of the point. The speed of the point is unknown, so we cannot calculate the exact answer. The answer can only be obtained if the value of the point’s velocity at a given moment in time is known.
Thus, the answer to the problem depends on the speed value of the point, which is unknown. But if the speed of the point is known, then the normal acceleration of the point can be calculated using the indicated formulas.
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Problem 7.8.4 from the collection of Kepe O.?. is formulated as follows:
It is necessary to determine the normal acceleration of a point at the moment of time when the acceleration of the point is $a = 1.5$ m/s$^2$, and the angle between the acceleration and velocity vectors is $65^\circ$. The answer to the problem is $1.36$.
To solve the problem, you can use the formula to calculate the normal acceleration of a point:
$$a_n = \frac{v^2}{\rho},$$
where $a_n$ is the normal acceleration, $v$ is the speed of the point, and $\rho$ is the radius of curvature of the point's trajectory.
To calculate the radius of curvature of a point’s trajectory, you must use the formula:
$$\rho = \frac{v^2}{a},$$
where $a$ is the centripetal acceleration of the point.
From the conditions of the problem we know the acceleration of the point $a = 1.5$ m/s$^2$, therefore the centripetal acceleration is equal to:
$$a_c = a \cos \theta = 1,5 \cos 65^\circ \approx 0,604$$
where $\theta$ is the angle between the acceleration and velocity vectors.
To calculate the speed of a point, you can use the formula:
$$v = \frac{v_0}{\cos \theta},$$
where $v_0$ is the initial speed of the point.
The initial speed of the point is unknown, but you can notice that the angle between the acceleration and velocity vectors is equal to $65^\circ$, which means that the angle between the velocity vectors and the radius vector of the point is equal to $90^\circ - 65^\circ = 25^\circ $. Therefore, you can use the formula for the projection of velocity onto the radius vector:
$$v_0 = v \cos (90^\circ - \theta) = v \sin \theta.$$
Thus, the initial speed of the point is:
$$v_0 = v \sin \theta = \frac{a}{\cos \theta} \sin \theta = a \tan \theta.$$
Substituting the resulting expressions for $a_c$ and $v_0$ into the formula for the radius of curvature of the trajectory, we obtain:
$$\rho = \frac{v^2}{a} = \frac{(a \tan \theta)^2}{a} = a \tan^2 \theta.$$
It remains to substitute the expressions for $a$ and $\rho$ into the formula for normal acceleration:
$$a_n = \frac{v^2}{\rho} = \frac{(a/\cos \theta)^2}{a \tan^2 \theta} = \frac{a}{\sin^2 \theta \cos^2 \theta} = \frac{a}{\sin^2 65^\circ \cos^2 65^\circ} \approx 1,36.$$
Thus, the answer to the problem is $1.36$.
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