- #1
myko
- 13
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[>A heavy sphere of radius r = 1.00 meter is fixed with respect to the ground. A small uniform solid sphere is placed at the top of the larger sphere. After a slight disturbance, the smaller sphere begins to roll downward without slipping. How high h is the small sphere above the ground at the instant it loses contact with the large sphere
My atempt of solution:
I think, that at the moment when it loses contact, speed of the small sphere is higher than the one in $\Sigma F=mg\cos\theta=mv^2/r$. $\theta r$ is an arc traveled on the sphere. Also since it falls without slipping, $v=\omega r$, where $\omega$ is the angular velocity of the small sphere. The speed at a height h can be found from $$mg(2r)=1/2mv^2+mgh.$$
From foce equation I would get $$v=\sqrt{rg\cos\theta}$$
And from energy conservation:
$$h=\frac{4g-v^2}{2g}$$
substituting the expretion for velocity
$$h=\frac{4-r\cos\theta}{2}$$
I am not sure if this is correct or no. And I can't express $\theta$ in some other way. Could anyone help?
My atempt of solution:
I think, that at the moment when it loses contact, speed of the small sphere is higher than the one in $\Sigma F=mg\cos\theta=mv^2/r$. $\theta r$ is an arc traveled on the sphere. Also since it falls without slipping, $v=\omega r$, where $\omega$ is the angular velocity of the small sphere. The speed at a height h can be found from $$mg(2r)=1/2mv^2+mgh.$$
From foce equation I would get $$v=\sqrt{rg\cos\theta}$$
And from energy conservation:
$$h=\frac{4g-v^2}{2g}$$
substituting the expretion for velocity
$$h=\frac{4-r\cos\theta}{2}$$
I am not sure if this is correct or no. And I can't express $\theta$ in some other way. Could anyone help?