Question

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 scale. 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.

Answer 1

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

Answer 2

At its lowest point, the net force of the bowling ball is equal to its net weight and this is given by $F_{net} = mg + \frac{2mgh}{r}$

Given the following data:

• Mass, m = 1.7 kg.

• Height, h = 14.2 meters.

• Radius, r = 6.2 meters.

To write an expression for the reading of the scale when the bowling ball is at its lowest point, in terms of the given variables:

The forces acting on the ball.

In this scenario, there are two (2) forces acting on the bowling ball and these include:

• Centripetal force

• Gravitational force

Mathematically, centripetal force is given by this formula:

$F_c = \frac{mv^2}{r}$     .....equation 1.

Mathematically, gravitational force is given by this formula:

$F_g= mg$    ....equation 2.

Where:

• m is the mass.

• g is the acceleration due to gravity.

• v is the velocity.

Next, we would derive an expression for the velocity of the ball by applying the Law of Conservation of energy:

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\V^2=2gh$ .....equation 3.

Substituting eqn. 3 into eqn. 2, we have:

$F_c = \frac{m(2gh)}{r}\\\\F_c = \frac{2mgh}{r}$

At its lowest point, the net weight of the bowling ball is equal to its net force and this is given by this mathematical expression:

$W_{net} = F_{net} = F_g + F_c\\\\F_{net} = mg + \frac{2mgh}{r}$

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