Questionthe ExpertTA.com Student: cristian.alvarez@ctstate.edu My Account Class Management I Help EXPERT Problem Status
HW7 Angular Motion Begin Date: 11/4/2024 12:01:00 AM Due Date: 11/22/2024 11:59:00 PM End Date: 12/13/2024 11:59:00 PM Problem 4: ( $25 \%$ of Assignment Value) A bowling ball of mass $m=2.4 \mathrm{~kg}$ drops from a height $h=14.4 \mathrm{~m}$ . A semi-circular tube of radius $r=6.2 \mathrm{~m}$ rests centered on a scale. Alvarez, Cristian - cristian.alvarez@ctstate.edu @theexpertta.com - tracking id: 8C84-EB-49-43-A913-26821. In accordance with Expert TA's Terms of Service, copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. Ctheexpertta.com
Part (a) 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 and $g$ . $W=$ $\square$ g 7 8 9 HOME 4 5 6 $\square$ 1 2 3 $\square$ 0 . END Grade Summary Deductions Potential Late Work \% Late Potential $10 \% \%$ $100 \%$ $78 \%$ $78 \%$ $\mathbf{7 8 \%}$ h m Submissions Attempt(s) Remaining: $5 \%$ Deduction per Attempt detailed view r - backspace DiE clear

Studdy Solution

STEP 1

1. The bowling ball is in free fall and then enters a semi-circular tube.
2. The scale measures the normal force exerted by the ball at the lowest point of the tube.
3. The gravitational acceleration is $g$ .

STEP 2

1. Determine the velocity of the ball at the lowest point.
2. Calculate the centripetal force required at the lowest point.
3. Determine the normal force on the scale.
4. Write the expression for the scale reading.

STEP 3

Determine the velocity of the ball at the lowest point using energy conservation. The potential energy at the height $h$ is converted into kinetic energy at the lowest point.
Potential energy at height $h$ : $PE = mgh$
Kinetic energy at the lowest point: $KE = \frac{1}{2}mv^2$
Setting $PE = KE$ , we have: $mgh = \frac{1}{2}mv^2$
Solve for $v$ : $v = \sqrt{2gh}$

STEP 4

Calculate the centripetal force required at the lowest point. The centripetal force $F_c$ is given by:
$F_c = \frac{mv^2}{r}$
Substitute $v = \sqrt{2gh}$ into the centripetal force equation: $F_c = \frac{m(2gh)}{r} = \frac{2mgh}{r}$

STEP 5

Determine the normal force on the scale. At the lowest point, the normal force $N$ is the sum of the gravitational force and the centripetal force:
$N = mg + F_c$
Substitute $F_c = \frac{2mgh}{r}$ : $N = mg + \frac{2mgh}{r}$

STEP 6

Write the expression for the scale reading. The scale reads the normal force, so:
$W = mg + \frac{2mgh}{r}$
This is the expression for the reading of the scale when the bowling ball is at its lowest point.
The expression for the scale reading is $W = mg + \frac{2mgh}{r}$ .

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