Unit 7 UCM Review Flashcards

Fictitious Forces: A fictitious force is an apparent force that arises from the acceleration of an observer's frame of reference rather than from any physical interaction. In the context of Uniform Circular Motion (UCM), the most common fictitious force discussed is Centrifugal Force.
Centripetal Definition: The term "centripetal" literally means "center seeking."
Acceleration in UCM: Even if an object moves at a constant speed, it is still undergoing acceleration if it is in a circular path because it is constantly changing direction as it moves around the circle.

Case Study: Car in a Valley: For a car with a mass of $1,000\,kg$ traveling at a speed of $25\,m/s$ through a valley with a radius of curvature ( $r$ ) of $25\,meters$ , the centripetal force ( $F_c$ ) is calculated as: - $F_c = \frac{m \times v^2}{r}$ - $F_c = \frac{1,000\,kg \times (25\,m/s)^2}{25\,m}$ - $F_c = 25,000\,N$
Case Study: Ball on a String: A $2.0\,kg$ ball is tied to a string and swung in a horizontal circle at a constant speed of $8.0\,m/s$ with a radius of $1.5\,meters$ . The force acting on the string is: - Given: $m = 2.0\,kg$ , $v = 8.0\,m/s$ , $r = 1.5\,m$ - $F_c = \frac{2 \times 8^2}{1.5}$ - $F_c = 85.3\,N$
Case Study: Calculating Speed from Force: A $1.2\,kg$ object moves in a circular path with a radius of $2.5\,meters$ and experiences a centripetal force of $18.0\,N$ . To find the speed ( $v$ ): - $18 = \frac{1.2 \times v^2}{2.5}$ - $18 \times 2.5 = 1.2 \times v^2$ - $45 = 1.2 \times v^2$ - $v = \sqrt{\frac{45}{1.2}}$ - $v = 6.12\,m/s$

The Gravitron: On this ride, it appears that riders are being pushed outward against the wall (centrifugal sensation). In reality, the walls are being pulled toward the center by a centripetal force, and the walls consequently push the riders inward to maintain the circular path.
Mario's Pizza Dough Calculations: - Frequency ( $f$ ) = $60\,revolutions/minute$ . - Period ( $T$ ): The time for one revolution is calculated as $T = \frac{1}{f}$ . In this case, $T = \frac{1}{60} = 0.016\,seconds$ (Note: this calculation assumes revolutions per second based on the specific work shown). - Tangential Velocity ( $v$ ) of pepperoni stuck $0.10\,meters$ from the center: - $v = \frac{2 \times \pi \times r}{T}$ - $v = \frac{2 \times \pi \times 0.1}{0.016}$ - $v = 39.27\,m/s$ - Centripetal Acceleration ( $a_c$ ): - $a_c = \frac{v^2}{r}$ - $a_c = \frac{(39.27)^2}{0.1}$ - $a_c = 15,421.39\,m/s^2$

Direction of Forces and Acceleration in a Loop: - Loop Bottom: The direction of acceleration and the net force is directed up. - Loop Top: The direction of acceleration and the net force is directed down.
Magnitude Comparisons: - Loop Bottom: The normal force is greater than the gravity force (F_{norm} > F_{grav}). - Loop Top: The normal force is less than the gravity force (F_{norm} < F_{grav}).
Roller Coaster Hill Scenarios ( $600-kg$ car, $F_{grav} = 5,800\,N$ ): - Hill A ( $v = 6\,m/s, R = 8\,m$ ): - $a_c = \frac{6^2}{8} = 4.5\,m/s^2$ - $F_c = 600 \times 4.5 = 2,700\,N$ - $F_{norm} = 5,800 - 2,700 = 3,100\,N$ - Hill B ( $v = 5\,m/s, R = 6\,m$ ): - $a_c = \frac{5^2}{6} = 4.17\,m/s^2$ - $F_c = 4.17 \times 600 = 2,502\,N$ - $F_{norm} = 5,800 - 2,502 = 3,298\,N$ - Hill C ( $v = 4\,m/s, R = 4\,m$ ): - $a_c = \frac{4^2}{4} = 4\,m/s^2$ - $F_c = 4 \times 600 = 2,400\,N$ - $F_{norm} = 5,800 - 2,400 = 3,400\,N$

Apparent Weight in a Depression: An $80\,kg$ driver ( $F_g = 784\,N$ ) in a depression with a radius of $20\,meters$ wants to feel twice their normal body weight ( $F_N = 2 \times F_g = 1,568\,N$ ). - The net centripetal force is $F_c = F_N - F_g = 1,568 - 784 = 784\,N$ . - Note: The transcript data shows a slight variation in subtraction: $F_c = 1,568 + (-789) = 779\,N$ . - Using $F_c = 779\,N$ : - $779 = \frac{80 \times v^2}{20}$ - $15,580 = 80 \times v^2$ - $v = \sqrt{\frac{15,580}{80}} = 13.95\,m/s$

Noah Formula's Golf Ball: When a golf ball leaves the outside rim of a circular green, it follows a straight-line path (Path 2) because if there is a break in the centripetal force, there is nothing left to keep it turning. The ball will travel straight, tangential to the circle and perpendicular to the radius.

Proportional Relationships: - When comparing sphere A and B (where B has a larger radius but the same period $T$ ), the speed of A is less than B (A < B). - The centripetal force on A is less than B (A < B) because $v$ and $r$ have a direct relationship when $T$ is constant.
Emma on the Spinning Ride ( $m = 60.0\,kg, r = 4.0\,m, F_c = 540\,N$ ): - $540 = \frac{60 \times v^2}{4}$ - $2,160 = 60 \times v^2$ - $v^2 = 36$ - $v = 6\,m/s$
Rotational Frequency and Period (for 50 rotations in 20 seconds): - Period ( $T$ ): T = \frac{\text{total time}}{\text{# of rotations}} = \frac{20}{50} = 0.4\,seconds - Frequency ( $f$ ): f = \frac{\text{# of rotations}}{\text{time}} = \frac{50}{20} = 2.5\,Hz
Inverse Relationships: - As radius decreases, acceleration increases (inverse relationship). - If an object takes three times as long ( $3 \times T$ ) to revolve at the same radius, the tangential velocity becomes one-third ( $1/3$ ) of the original velocity ( $v = \frac{2\pi r}{T}$ ).