Question

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.

Answer 1

• Coefficient of static friction = 0.5
• Coefficient of Kinetic friction = 0.3
• Angular velocity = 500 RPMs

The Radius of the System

Let R be the radius of cylinder

$m_a + m_b = 4 + 3 = 7kg$

The angular velocity is 500 RPMs

$\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R$

The normal force

$f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm$

Since the radius is very little for two block to execute circular motion so system will slide down.

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