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kvasek [131]
3 years ago
5

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070

0 kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?
Physics
1 answer:
likoan [24]3 years ago
8 0

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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enyata [817]

Answer:

Explanation:

Given:

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See the explanation below.

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<u>Explanation:</u>

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mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2

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Applying the principle of Impulse and momentum in y-direction

mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2

Limit is t1 to t2

(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}                            -(2)

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8 0
3 years ago
• ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.
nignag [31]

Answer:

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Explanation:

Let's take the north-south direction as y-direction (with south being positive) and east-west direction as x-direction (with west being positive). Therefore, the two components of Cody's motion are:

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Since they are perpendicular, the magnitude of the net displacement can be calculated by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{7.50^2+45.0^2}=45.6 m

The direction instead can be measured as follows:

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maks197457 [2]

Answer:

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