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3 years ago

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A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left and two weights of the same mass, on the right.

The width of each weight is 4 cm and each group of weights is placed 4 cm from the ends. Where is the center of mass of the barbell as measured from the mid-point, M, of the bar? The bar is of uniform mass and has mass 5 kg, and the retaining collars are of negligible mass. Take to the right as positivea. -5.90cmb. -11.6cmc. +13.7cmd. +5.90cme. none of the above

1 answer:

VMariaS [17]3 years ago

4 0

Answer:

b. -11.6 cm

Explanation:

We have given parameters:

Length, l = 1.5 m = 150 cm

Mass of weight, $m_1$ = 20 kg

Width, x = 4 cm

Distance d = 4 cm

Mass of bar, $m_{bar}$ = 5 kg

We are asked to find the center of mass from the mid-point, $X_{CM} = ?$

Since 3 weights are on the left and 2 weights are on the right, we know:

$m_{left}$ = 3 * 20 = 60 kg

$m_{right}$ = 2 * 20 = 40 kg

And also we know that, $M = \frac{l}{2}$ = 150/2 = 75 cm

For the left side, center of mass is:

$x_{left} = \frac{3 * 4}{2} = 6$ cm

From the midpoint, the distance to the left is:

$X_{left} = -(M - 4 - x_{left}) = -(75 - 4 -6) = -65$ cm

For the right side, center of mass is:

$x_{right} = \frac{2 * 4}{2} = 4$ cm

From the midpoint, the distance to the right will be:

$X_{right} = (M - 4 - x_{right}) = (75 - 4 - 4) = 67$ cm

Hence,

$X_{CM} = \frac{m_{right}*x_{right} + m_{left}*x_{left} }{m_{right} + m_{left} + m_{bar}} = \frac{40 * 67 - 60 * 65}{40 + 60 + 5} = -11.62$ cm

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The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

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Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

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therefore the acceleration is

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We can find the duration of the collision in both cases by using the equation of the acceleration

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For the hardwood floor,

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So the duration of the collision is

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"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

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