Question

# There is a clever kitchen gadget for drying lettuce leavesafter you wash them. It consists of a cylindrical container mountedso that it can be rotated

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There is a clever kitchen gadget for drying lettuce leavesafter you wash them. It consists of a cylindrical container mountedso that it can be rotated about its axis by turning a hand crank.The outer wall of the cylinder is perforated with small holes. Youput the wet leaves in the container and turn the crank to spin offthe water. The radius of the container is 13 cm. When the cylinder is rotating at1.6 revolutions per second, what isthe magnitude of the centripetal acceleration at the outerwall?
$$m/s^{2}$$

2021-01-06

1. The centripetal acceleration is
$$a = r_2 = (0.14)(2\cdot1.7)2 =15.9729\ m/s^2$$
2. Given that
The radius $$(r)=10cm=0.10m$$
The angular valocity is revolutions/sec
The centripetal acceleration is substitute the above given values you get the required answer
3. Radius of the container $$R =14cm$$
Angular velocity of the cylinder $$w= 1.7rev/s$$
Centripetal accelearation $$a = Rw^{2}$$
4. The equation for centripetal acceleration is $$a = v^2/R$$; where v is velocity and R is radius. The equation for velocity is change in X (position) divided by the change in time, t. The change of position during a circle motion is $$2(PI)R \times \#$$ of Revolutions all divided by time, t. So $$v = \frac{2(3.14)(0.11m)(2.4)}{1sec}$$ Place this velocity in the equation for centripetal acceleration $$(v^2/R)$$ and you should get an answer of $$[6.25\ m/sec^2]$$.