Using Circular Motion Equations - High School Physics

Card 0 of 20

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the centripetal acceleration?

Answer

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$a_c=\frac{9\frac{m^2}{s^2}}{1m}$

$a_c=9\frac{m}{s^2}$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the centripetal force?

Answer

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is .

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$a_c=\frac{9\frac{m^2}{s^2}}{1m}$

$a_c=9\frac{m}{s^2}$

Plug that into the first equation to solve for the force.

$F_c=(2kg)(9\frac{m}{s^2})$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the period of the ball?

Answer

The period, , of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.

If we start with the linear understanding of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply the same concept here. Our velocity should be the change in distance over the change in time. In this case, we don't have a definite time, , but we do have a period in terms of one complete loop.

We can set up an equation for the period using the circumference of the circle as our distance: $v=\frac{2\pi r}{T}$ .

Plug in the given values to solve for the period.

$v=\frac{2\pi r}{T}$

$3\frac{m}{s}=\frac{2\pi (1m)}{T}$

$T=\frac{2\pi (1m)}{3\frac{m}{s}}$

$T=\frac{2\pi}{3}s$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $8\frac{m}{s}$ , what is the centripetal acceleration?

Answer

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$a_c=128\frac{m}{s^2}$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $8\frac{m}{s}$ , what is the centripetal force?

Answer

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is .

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$a_c=128\frac{m}{s^2}$

Plug the acceleration and given mass into the first equation to solve for force.

$F_c=(2kg)(128\frac{m}{s^2})$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $8\frac{m}{s}$ , what is the period of the ball?

Answer

The period, , of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.

If we start with the linear understanding of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply the same concept here. Our velocity should be the change in distance over the change in time. In this case, we don't have a definite time, , but we do have a period in terms of one complete loop.

We can set up an equation for the period using the circumference of the circle as our distance: $v=\frac{2\pi r}{T}$ .

Plug in the given values to solve for the period.

$v=\frac{2\pi r}{T}$

$8\frac{m}{s}=\frac{2\pi (0.5m)}{T}$

$T=\frac{\pi m}{8\frac{m}{s}}$

$T=\frac{\pi}{8}s$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If there is a centripital force acting upon it, what is its speed?

Answer

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is .

Since we know the mass and the force, we can find the accleration.

$\frac{12N}{2kg}=a_c$

$6\frac{m}{s^2}=a_c$

Centripetal acceleration is given by the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the velocity.

$6\frac{m}{s^2}=\frac{v^2}{(1m)}$

$1m* 6\frac{m}{s^2}=v^2$

$6\frac{m^2}{s^2}=v^2$

$\sqrt{6\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.45\frac{m}{s}=v$

Compare your answer with the correct one above

Question

A ball rolls around the edge of a circle with a radius of . If it has a centripial force of acting upon it, what is its speed?

Answer

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is .

Since we know the mass and the force, we can find the accleration.

$\frac{8N}{2kg}=a_c$

$4\frac{m}{s^2}=a_c$

Centripetal acceleration is given by the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the velocity.

$4\frac{m}{s^2}=\frac{v^2}{(1m)}$

$1m*4\frac{m}{s^2}=v^2$

$4\frac{m^2}{s^2}=v^2$

$\sqrt{4\frac{m^2}{s^2}}=\sqrt{v^2}$

$2\frac{m}{s}=v$

Compare your answer with the correct one above

Question

The radius of the sun is 696,000km. If its period is 587.28 hours, what is its tangential velocity at the equator?

Answer

The equation for velocity is $v=\frac{\Delta x}{\Delta t}$ . Using the circumference of the circle as the distance and the time as the period, we can rewrite the equation for velocity: $v=\frac{2\pi r}{T}$ .

Plug in the given values and solve for the velocity.

$v=\frac{2\pi (696,000km)}{587.28hr}$

$v=\frac{\pi*1392000km}{587.28hr}$

$v=2370.25\pi\frac{km}{hr}$

Compare your answer with the correct one above

Question

An object moves in a circle with a constant velocity of $3\frac{m}{s}$ . If the radius of the circle is , what is the centripetal acceleration on the object?

Answer

The formula for centripetal acceleration is:

$a_c=\frac{v^2}{r}$

We are given the velocity and the radius, allowing us to solve for the acceleration.

$a_c=\frac{v^2}{r}$

$a_c=\frac{(3\frac{m}{s})^2}{2m}$

$a_c=\frac{9\frac{m^2}{s^2}}{2m}$

$a_c=4.5\frac{m}{s^2}$

Compare your answer with the correct one above

Question

If the centripetal force on a object is , and the object is from the center of the circle, what is its centripetal acceleration?

Answer

The formula for centripetal force is .

We know the value of the force, as well as the mass of the object. Using these values, we can solve for the acceleration; the radius is extraneous information.

$\frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$

Compare your answer with the correct one above

Question

If the centripetal force on a object is , and the object is from the center of the circle, what is its tangential velocity?

Answer

The relationship between centripetal acceleration and tangential velocity is:

$a_c=\frac{v^2}{r}$

We can find the centripetal acceleration using the centripetal force and the mass.

The formula for centripetal force is .

Using the force and mass from the question, we can solve for the acceleration.

$\frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$

Now that we know the acceleration, we can return to the first equation. Using the acceleration and the radius from the question, we can solve for the velocity.

$a_c=\frac{v^2}{r}$

$4.34\frac{m}{s^2}=\frac{v^2}{1.9m}$

$4.34\frac{m}{s^2}*1.9m=v^2$

$8.246\frac{m^2}{s^2}=v^2$

$\sqrt{8.246\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.87\frac{m}{s}=v$

Compare your answer with the correct one above

Question

A large burst of wind exerts of force on a tree tall. How much torque is exerted on the tree? (Assume that the entire force is pointed at the top of the tree, the pivot point is at the ground, and ignore center of mass fluctuations)

Answer

The formula for torque is:

$\tau=Fr$

Plug in our given values:

$\tau=750N10m$

$\tau=7500N*m$

Compare your answer with the correct one above

Question

of force is exerted on the end of a wrench. How much torque is generated? (Ignore any center of mass fluctuations.)

Answer

The formula for torque is:

$\tau=Fr$

Plug in our given values:

$\tau=8N0.23m$

$\tau=1.84N*m$

Compare your answer with the correct one above

Question

The velocity of a certain ferris wheel is $6\frac{m}{s}$ . If the wheel has a diameter of , what is the centripetal acceleration generated by the wheel?

Answer

The formula for centripetal acceleration is $a_c=\frac{v^2}{r}$ .

The problem gives us a diamater. Since radius is defined as half of an object's diameter, the radius of the ferris wheel is .

Now we can plug in and solve:

$a_c=\frac{v^2}{r}$

$a_c=\frac{(6\frac{m}{s})^2}{6m}$

$a_c=\frac{36\frac{m^2}{s^2}}{6m}$

$a_c=6\frac{m}{s^2}$

Compare your answer with the correct one above

Question

The torque applied to a wrench is . If the length of the wrench is , how much force is applied to it?

Answer

The formula for torque is:

$\tau=Fr$

We are given the total torque and the length of the wrench. Given that the pivot point will be at one end of the wrench, and the force will be applied to the other end, this length can be used as our radius.

$\frac{14Nm}{0.5m}=F$

Compare your answer with the correct one above

Question

The torque applied to a wrench is . If the force applied to the wrench is , how long is the wrench?

Answer

The formula for torque is:

$\tau=Fr$

We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.

$\frac{16Nm}{40N}=r$

Compare your answer with the correct one above

Question

If the force applied to a wrench is , how much torque is generated?

Answer

The formula for torque is:

$\tau=Fr$

We are given the values for the force and the length. Using these values, we can multiply to find the torque.

$\tau=50N0.21m$

$\tau=10.5N*m$

Compare your answer with the correct one above

Question

How much force is required for a hammer to produce of torque?

Answer

The formula for torque is:

$\tau=Fr$

We are given the length of the hammer (radius of the swing) and the torque produced. Using these values, we can solve for the force required.

$\frac{4.2Nm}{0.3m}=F$

Compare your answer with the correct one above

Question

A car makes a right turn. The radius of this curve is . If the force of friction between the tires and the road is , what is the maximum velocity that the car can have before skidding?

Answer

To solve this problem, recognize that the force due to friction must equal the centripetal force of the curve:

This will give the maximum force that the car can have in the curve without skidding. Expand the equation for centripetal force.

$F_f=m\frac{v^2}{r}$

We are given the value for the force of friction, the mass of the car, and the radius of the curve. Using these values, we can find the velocity.

$2587.2N=1200kg\frac{v^2}{4.2m}$

$\frac{2587.2N}{1200kg}=\frac{v^2}{4.2m}$

$2.156\frac{m}{s^2}=\frac{v^2}{4.2m}$

$4.2m2.156\frac{m}{s^2}=v^2$

$9.0552\frac{m^2}{s^2}=v^2$

$\sqrt{9.0552\frac{m^2}{s^2}}=\sqrt{v^2}$

$3.01\frac{m}{s}=v$

Compare your answer with the correct one above

Tap the card to reveal the answer