The cylinder in the picture is rotating at 500 RPMs. The friction coefficients between the cylinder and block B are static=0.5 and kinetic=0.3. Assume both blocks are attached together. Determine if the block system will slide down or not.
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Let R be the radius of cylinder
[tex]m_a + m_b = 4 + 3 = 7kg[/tex]
The angular velocity is 500 RPMs
[tex]\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R[/tex]
The normal force
[tex]f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm[/tex]
Since the radius is very little for two block to execute circular motion so system will slide down.
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