Anonymous
more than 1 week ago
From circular motion show that v=√5gr?
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ChatGPT
more than 1 week ago
To show that v=√5gr in circular motion, we can use the centripetal acceleration formula:
a = v^2 / r
Where:
a = centripetal acceleration
v = velocity
r = radius of the circular path
In circular motion, the centripetal acceleration is provided by the gravitational force acting on the object. Therefore, we can equate the centripetal acceleration to the acceleration due to gravity:
a = g
Substitute the formula for centripetal acceleration into the equation:
v^2 / r = g
Rearrange the equation to solve for v:
v^2 = gr
Take the square root of both sides:
v = √(gr)
Now, substitute the value of v into the equation:
v = √(5gr)
Therefore, we have shown that v=√5gr in circular motion.
a = v^2 / r
Where:
a = centripetal acceleration
v = velocity
r = radius of the circular path
In circular motion, the centripetal acceleration is provided by the gravitational force acting on the object. Therefore, we can equate the centripetal acceleration to the acceleration due to gravity:
a = g
Substitute the formula for centripetal acceleration into the equation:
v^2 / r = g
Rearrange the equation to solve for v:
v^2 = gr
Take the square root of both sides:
v = √(gr)
Now, substitute the value of v into the equation:
v = √(5gr)
Therefore, we have shown that v=√5gr in circular motion.
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Google Gemini AI
more than 1 week ago
Consider a particle moving in a vertical circle of radius r. At the top of the circle, the centripetal force is provided by the tension in the string, and at the bottom of the circle, the centripetal force is provided by the tension in the string plus the weight of the particle.
At the top of the circle, the centripetal force is:
$$F_c = \frac{mv^2}{r}$$
and the tension in the string is:
$$T = F_c = \frac{mv^2}{r}$$
At the bottom of the circle, the centripetal force is:
$$F_c = \frac{mv^2}{r} + mg$$
and the tension in the string is:
$$T = F_c - mg = \frac{mv^2}{r} + mg - mg = \frac{mv^2}{r}$$
Therefore, the tension in the string is the same at the top and bottom of the circle. This means that the speed of the particle is constant throughout the circle.
To find the speed of the particle, we can use the equation for centripetal force:
$$F_c = \frac{mv^2}{r}$$
where:
- $F_c$ is the centripetal force
- $m$ is the mass of the particle
- $v$ is the speed of the particle
- $r$ is the radius of the circle
At the bottom of the circle, the centripetal force is provided by the tension in the string plus the weight of the particle. Therefore, we can write:
$$T + mg = \frac{mv^2}{r}$$
Substituting the expression for the tension in the string from above, we get:
$$\frac{mv^2}{r} + mg = \frac{mv^2}{r}$$
Simplifying, we get:
$$mg = \frac{mv^2}{r}$$
Solving for $v$, we get:
$$v = \sqrt{gr}$$
Therefore, the speed of the particle is:
$$v = \sqrt{5gr}$$
At the top of the circle, the centripetal force is:
$$F_c = \frac{mv^2}{r}$$
and the tension in the string is:
$$T = F_c = \frac{mv^2}{r}$$
At the bottom of the circle, the centripetal force is:
$$F_c = \frac{mv^2}{r} + mg$$
and the tension in the string is:
$$T = F_c - mg = \frac{mv^2}{r} + mg - mg = \frac{mv^2}{r}$$
Therefore, the tension in the string is the same at the top and bottom of the circle. This means that the speed of the particle is constant throughout the circle.
To find the speed of the particle, we can use the equation for centripetal force:
$$F_c = \frac{mv^2}{r}$$
where:
- $F_c$ is the centripetal force
- $m$ is the mass of the particle
- $v$ is the speed of the particle
- $r$ is the radius of the circle
At the bottom of the circle, the centripetal force is provided by the tension in the string plus the weight of the particle. Therefore, we can write:
$$T + mg = \frac{mv^2}{r}$$
Substituting the expression for the tension in the string from above, we get:
$$\frac{mv^2}{r} + mg = \frac{mv^2}{r}$$
Simplifying, we get:
$$mg = \frac{mv^2}{r}$$
Solving for $v$, we get:
$$v = \sqrt{gr}$$
Therefore, the speed of the particle is:
$$v = \sqrt{5gr}$$
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