اختبار اختيار من متعدد تفاعلي: Circular Motion Worksheets - inspire
Circular motion describes the movement of an object along the circumference of a circle or a curved path. It is defined by several key physical quantities, most notably centripetal acceleration and centripetal force. Even when an object moves at a constant speed, it undergoes acceleration because its direction of travel is constantly changing. This centripetal acceleration is always directed toward the center of the circular path and can be calculated using the formula \( a_c = \frac{v^2}{r} \), where v is the tangential speed and r is the radius. Consequently, a centripetal force defined by \( F_c = \frac{mv^2}{r} \) must act on the object to maintain this motion, where m is the mass. This worksheet tests the understanding of these relationships, units, and vector directions in the context of uniform circular motion.
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اختبار شهادة تدريبي مؤقت للصف والمادة والفصل نفسه.
اختر إجابة واحدة لكل سؤال. عند الاختيار ستظهر النتيجة فورًا: الأخضر صحيح، والأحمر خطأ، وسيظهر تفسير الإجابة مباشرة إن كان متوفرًا. وبعد آخر سؤال ستظهر الدرجة النهائية تلقائيًا.
A pilot experiences centripetal acceleration of 2g's (2 × 9.8 m/s2) while maneuvering a loop at a speed of 400 m/s. What is the radius of the loop?
Explanation
Using the centripetal acceleration formula \( a_c = \frac{v^2}{r} \), we rearrange for radius: \( r = \frac{v^2}{a_c} \). Substituting the values, \( r = \frac{400^2}{19.6} = \frac{160000}{19.6} \approx 8163 \text{ m} \).
Question 2
Points: 1
What is the SI unit for circumference?
Explanation
Circumference is a measure of length (the distance around a circle). The International System of Units (SI) unit for length is the meter.
Question 3
Points: 1
What happens to the centripetal force exerted on an object if you triple the velocity?
Explanation
Centripetal force is proportional to the square of the velocity \( (F_c \propto v^2) \). If velocity is tripled (3v), the force increases by a factor of 32 = 9.
Question 4
Points: 1
A satellite of mass m and speed v orbits the Earth at a distance r from the center of the Earth. The gravitational acceleration due to the Earth at the satellite is equal to:
Explanation
In a circular orbit, the gravitational acceleration provides the centripetal acceleration, which is defined as \( a_c = \frac{v^2}{r} \).
Question 5
Points: 1
Calculate the velocity of an object moved around a circle with a radius of 1.65 m and an acceleration of 3.5 m/s2.
Explanation
Using \( a_c = \frac{v^2}{r} \), we solve for velocity: \( v = \sqrt{a_c \cdot r} = \sqrt{3.5 \times 1.65} = \sqrt{5.775} \approx 2.4 \text{ m/s} \).
Question 6
Points: 1
Calculate the centripetal force required to make a 5 kg ball move 4 m/s in a circle with a radius of 2 m.
Explanation
Using the formula \( F_c = \frac{mv^2}{r} \): \( F_c = \frac{5 \times 4^2}{2} = \frac{5 \times 16}{2} = 40 \text{ N} \).
Question 7
Points: 1
Centripetal force is...
Explanation
By definition, 'centripetal' means center-seeking. It is the net force that acts toward the center of the circle to maintain circular motion.
Question 8
Points: 1
Centripetal acceleration acts towards the ___________ of the circular motion
Explanation
Centripetal acceleration always points toward the center of the circular path of an object.
Question 9
Points: 1
A horse gallops around a circular arena. The centripetal force on the horse is ___.
Explanation
In any circular motion, the centripetal force is always directed toward the center of the curvature/circle.
Question 10
Points: 1
Calculate the mass of an object if it took 20 N to rotate it in a circle with a radius of 2 meters with a velocity of 4 m/s
Explanation
From the formula \( F_c = \frac{mv^2}{r} \), we solve for mass: \( m = \frac{F_c \cdot r}{v^2} = \frac{20 \times 2}{4^2} = \frac{40}{16} = 2.5 \text{ kg} \).
Question 11
Points: 1
Why do objects experience centripetal acceleration even though they can have a constant speed?
Explanation
Acceleration is defined as the rate of change of velocity. Since velocity is a vector, a change in direction constitutes a change in velocity, even if the speed (magnitude) remains constant.
Question 12
Points: 1
A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track's center and has a centripetal acceleration of 8.05 m/s2, what is the car's tangential speed?
Explanation
Using \( v = \sqrt{a_c \cdot r} \): \( v = \sqrt{8.05 \times 48.2} = \sqrt{388.01} \approx 19.7 \text{ m/s} \).
Question 13
Points: 1
The linear velocity is always _____ to the line of a circle.
Explanation
Linear or tangential velocity is the instantaneous direction of motion, which is always tangent to the circular path at any point.
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