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snow_lady [41]
3 years ago
14

A bowling ball of mass m = 1.7 kg drops from a height h = 14.2 m. A semi-circular tube of radius r = 6.2 m rest centered on a sc

ale. Write an expression for the reading of the scale when the bowling ball is at its lowest point, in terms of the variables in the problem statement.
Physics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

W_net = mg + 2mgh/r

Explanation:

The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.

Now, gravitational force is; F_g = mg

While centripetal force is; F_c = mv²/r

Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.

Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.

Thus;

F_c = 2mgh/r

Where;

m is ball mass

r is tube radius

h is fall height

Thus, the net force will be;

F_net = F_g + F_c

Now, Net force would be equal to the net weight that will be read on the scale.

Thus;

W_net = F_net = F_g + F_c

W_net = mg + 2mgh/r

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The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

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