Chapter 7: Q. 56 (page 181)

A $40 c m$ diameter, $50 c m$ tall, $15 kg$ hollow cylinder is placed on top of a $40 c m$ -diameter, $30 c m$ -tall, $100 kg$ cylinder of solid aluminum, then the two are sent sliding across frictionless ice. The static and kinetic coefficients of friction between the cylinders are $0.45$ and $0.25$ , respectively. Air resistance cannot be neglected. What is the maximum speed the cylinders can have without the top cylinder sliding off?

Short Answer

Expert verified

The maximum speed the cylinders can have without the top cylinder sliding off is 34 m/s.

Step by step solution

Given.

Mass of the top cylinder $m_{1} = 15 Kg$

Mass of bottom cylinder $m_{2} = 100 Kg$

Length of the top cylinder $L_{1} = 50 cm = 0.5 m$

Length of the bottom cylinder $L_{2} = 30 cm = 0.3 m$

Diameter of the top cylinder $d = 40 cm = 0.4 m$

Diameter of the bottom cylinder $d = 40 cm = 0.4 m$

The static friction coefficient between them $= 0.45$

The kinetic friction coefficient between them $= 0.25$

To determine the speed of the cylinders in order that the upper cylinder won't slip. For determining this we want to contemplate that the difference in drag forces on both the cylinder should not exceed the static friction between the cylinders as air resistance can't be neglected.

If drag force is high it'll push the upper cylinder as a result the cylinder will slip.

Also because the cylinders need to be moving together hence their velocities is the same: $v_{1} = v_{2}$ Keeping this in mind we'll proceed as follow:

From diagram below

Static friction between them.

$f_{s} = μ_{s} n n = m_{1} g f_{s} = μ_{s} m_{1} g F_{d r a g 1} = \frac{1}{2} ρ C A_{1} {v_{1}}^{2} C = 1.1 A_{!} = 2 r_{1} L_{1} = 2 \times 0.2 \times 0.5 = 0.2 m^{2} F_{d r a g 1} = \frac{1}{2} \times 1.3 \times 1.1 \times 0.2 v^{2} = 0.143 v^{2} F_{d r a g 2} = \frac{1}{2} ρ C A_{2} {v_{2}}^{2} A_{!} = 2 r_{2} L_{2} = 2 \times 0.2 \times 0.3 = 0.12 m^{2} F_{d r a g 2} = \frac{1}{2} \times 1.3 \times 1.1 \times 0.12 v^{2} = 0.0858 v^{2} F_{d r a g 1} - F_{d r a g 2} = μ_{s} m_{1} g (0.143 - 0.0858) v^{2} = 0.45 \times 15 \times 9.8 v = \sqrt{\frac{66.15}{0.0572}} = 34 m / s$