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.
رقم الاختبار799
الصفالصف الحادي عشر المتقدم
المادةفيزياء
الفصلالفصل الثالث
السنة الدراسية2025/2026
عدد الأسئلة13
إجمالي النقاط13
تاريخ الإضافة2026-04-19
الزيارات149
الناشرAmal Salman
يرجى الانتباه إلى أن المعلم قام بإعداد الأسئلة فقط، ولم يقم بإعداد الإجابات أو الشروحات المرفقة. وقد تم توليد الإجابات باستخدام تقنيات الذكاء الاصطناعي، لذلك قد تتضمن بعض الأخطاء أو عدم الدقة.
للحصول على الإجابات الصحيحة والمضمونة، يُرجى الرجوع إلى المعلم أو المصدر الدراسي المعتمد.
Question 1
Points: 1
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?
Using the centripetal acceleration formula \( a_c = \frac{v^2}{r} \), we rearrange for radius: \( r = \frac{v^2}{a_c} \). Substituting the values, \( r = \frac{400^2}{19.6} = \frac{160000}{19.6} \approx 8163 \text{ m} \).
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:
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?
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?
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