AP Physics C: Mechanics Practice Quiz: Circular Motion

Correct Answer: C

For an object to move in a circular path, the direction of its velocity vector must constantly change, even if its speed (the magnitude of the velocity) is constant. This change in velocity implies there is an acceleration.

Correct Answer: C

Centripetal acceleration is given by the equation $a_{c}=rac{v^{2}}{r}$. Since acceleration is proportional to the square of the speed ($a_c \propto v^2$), doubling the speed (v → 2v) will result in the acceleration being multiplied by a factor of (2)^2 = 4.

Correct Answer: A

Centripetal acceleration is given by the equation $a_{c}=rac{v^{2}}{r}$. Since acceleration is inversely proportional to the radius ($a_c \propto 1/r$), doubling the radius (r → 2r) while keeping the speed constant will cause the acceleration to be halved.

Correct Answer: B

The equation $T^{2}=rac{4\pi^{2}}{GM}R^{3}$ shows that $T^2$ is directly proportional to $R^3$. The term $rac{4\pi^{2}}{GM}$ is the constant of proportionality for a given central body.

Correct Answer: C

From Kepler's third law, $T^2 \propto R^3$, which means $T \propto R^{3/2}$. The ratio of the periods is $rac{T_B}{T_A} = (rac{R_B}{R_A})^{3/2}$. Since $R_B = 4R_A$, the ratio is $(rac{4R_A}{R_A})^{3/2} = (4)^{3/2} = (\sqrt{4})^3 = 2^3 = 8$.

Correct Answer: A

The relationship is $T^{2}=rac{4\pi^{2}}{GM}R^{3}$. This shows that $T^2$ is inversely proportional to M ($T^2 \propto 1/M$), so $T \propto 1/\sqrt{M}$. If M is doubled (M → 2M), the new period $T'$ will be $T' \propto 1/\sqrt{2M}$, which means $T' = T/\sqrt{2}$. The period decreases by a factor of $\sqrt{2}$.

Correct Answer: D

The provided content explicitly states that the magnitude of centripetal acceleration is the ratio of the object's tangential speed squared to the radius of the circular path, which is mathematically expressed as $a_{c}=rac{v^{2}}{r}$.

Correct Answer: C

The original acceleration is $a_c = rac{v^2}{r}$. The new speed is $v' = 2v$ and the new radius is $r' = r/2$. The new acceleration $a'_c$ is $a'_c = rac{(v')^2}{r'} = rac{(2v)^2}{r/2} = rac{4v^2}{r/2} = 8 rac{v^2}{r} = 8a_c$.

Correct Answer: D

Analyzing the equation $T^{2}=rac{4\pi^{2}}{GM}R^{3}$, the variables are the period (T), the gravitational constant (G), the mass of the central body (M), and the orbital radius (R). The mass of the object that is in orbit is not present in this equation, so the period is independent of it.

Correct Answer: D

Centripetal acceleration is $a_{c}=rac{v^{2}}{r}$. To change $a_c$ to $4a_c$ while keeping $v$ constant, we have $4a_c = rac{v^2}{r'}$. Substituting $a_c$, we get $4(rac{v^2}{r}) = rac{v^2}{r'}$. This simplifies to $4/r = 1/r'$, so $r' = r/4$. The radius must be divided by 4.

Correct Answer: B

From Kepler's third law, $T^2 \propto R^3$, which can be rearranged to $R^3 \propto T^2$, or $R \propto T^{2/3}$. If the new period $T'$ is $27T$, the new radius $R'$ will be related by $R' \propto (27T)^{2/3} = (27^{2/3})T^{2/3}$. Since $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$, the new radius is 9 times the original radius.

Correct Answer: C

Centripetal acceleration is the acceleration required to continuously change the direction of the velocity vector to keep an object in a circular path. This acceleration is always directed radially inward, toward the center of the circle.

Correct Answer: B

We are given $a_c = rac{v^2}{r}$. We want the new acceleration $a'_c$ to be equal to $a_c$, with a new radius $r' = 4r$. So, $a_c = rac{(v')^2}{r'} = rac{(v')^2}{4r}$. Setting the two expressions for $a_c$ equal: $rac{v^2}{r} = rac{(v')^2}{4r}$. This simplifies to $v^2 = rac{(v')^2}{4}$, or $(v')^2 = 4v^2$. Taking the square root of both sides gives $v' = 2v$. The speed must be doubled.