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@kabimbokaputo
My bf got tumblr now I can’t use it
I’m @casadepalermo aka @qqwwertty aka @kabimbokaputo
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Alla Kostromichova by Sølve Sundsbø in “The Girl from Atlantis” for Vogue Japan May 2010 wearing Alexander McQueen | Snake skin
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My bf got tumblr now I can’t use it
I’m @casadepalermo aka @qqwwertty aka @kabimbokaputo
I told him he was invading my privacy bc my online journal was public to the entire world just not him
I do kind of like this blog tho
Throckmorton problem
I have the textbook Throckmorton is from so I’m gonna write the problem here. It’s a worked example, but I’m not gonna do the parts that work it out because there’s too much math for me to format.
Example 7.4 Speed at the bottom of a vertical circle
Your cousin Throckmorton skateboards from rest down a curved, frictionless ramp. If we treat Throcky and his skateboard as a particle, he moves through a quarter-circle with radius (Fig. 7.9). Throcky and his skateboard have a total mass of 25.0 kg.
(a) Find his speed at the bottom of the ramp.
(b) Find the normal force that acts on him at the bottom of the curve.
Figure 7.9
(a) Throcky skateboarding down a frictionless circular ramp. The total mechanical energy is constant.
(b) Free-body diagrams for Throcky and his skateboard at various points on the ramp.
[Text within figure: At each point, the normal force acts perpendicular to the direction of Throcky’s displacement, so only the force of gravity (w) does the work on him.]
Solution
Identify: We can’t use the constant-acceleration equations of Chapter 2 because Throcky’s acceleration isn’t constant; the slope decreases as he descends. Instead, we’ll use the energy approach. Throcky moves along a circular arc, so we’ll also use what we learned about circular motion in Section 5.4.
Set up: The only forces on Throcky are his weight and the normal force n exerted by the ramp (Fig. 7.9b). Although n acts all along the path, it does zero work because n is perpendicular to Throcky’s displacement at every point. Hence Wother = 0 and mechanical energy is conserved. We take point 1 at the starting point and point 2 at the bottom of the ramp, and we let y = 0 be at the bottom of the ramp (Fig. 7.9a). We take the positive y-direction upward; then y1 = R and y2 = 0. Throcky starts at rest at the top, so v1 = 0. In part (a) our target variable is his speed v2 at the bottom; in part (b) the target variable is the magnitude n of the normal force at point 2. To find n, we’ll use Newton’s second law and the relation a = v2/R.
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