Centripetal force and acceleration from mass, speed and radius; rpm to rad/s and period; moment of inertia of nine shapes; torque from force and lever arm.
آخر تحديث أمثلة تم التحقق منها: 7
Keeping 1,000 kg on a 50 m circle at 20 m/s takes 8,000 N directed at the centre — an acceleration of 0.816 g.
Centripetal force is the inward pull that keeps a mass on a circular path: F = mv²/r, with the centripetal acceleration a = v²/r = ω²r pointing at the centre. The other modes convert a rotation rate between rpm, rad/s, hertz and period (ω = 2πf), give the moment of inertia I of nine standard shapes, and work out torque τ = rF sin θ and angular acceleration α = τ/I.
The default is a 1,000 kg car taking a 50 m radius curve at 20 m/s (72 km/h). The tyres must supply 8,000 N of sideways grip, an acceleration of 8 m/s² or 0.816 g. In torque mode, a 50 N push at 90° on a 0.3 m wrench gives 15 N·m.
Results assume uniform circular motion and rigid bodies of uniform density. Mass farther from the axis raises I, so a thin hoop has twice the moment of inertia of a solid disc with the same mass and radius.
مصدر التحقق: Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4
مصدر التحقق: Python 3.8 decimal: v = √(Fr/m) = √100
مصدر التحقق: Python 3.8 math: ω = 2π·3000/60, v = ωr
مصدر التحقق: Serway Table 10.2: I = 2/5 MR² = 0.4 × 5 × 0.04
F = mv²/r, with mass m in kilograms, speed v in metres per second and radius r in metres, giving newtons. Equivalent forms are F = mω²r and F = 4π²mr/T². Doubling the speed quadruples the force and doubling the radius halves it: a 1,000 kg car at 20 m/s on a 50 m curve needs 8,000 N, and at 40 m/s it would need 32,000 N.
Multiply rpm by 2π/60, about 0.10472. 3,000 rpm is 314.159 rad/s, or 50 revolutions per second (50 Hz), so one revolution takes 0.02 s. To go back, multiply rad/s by 60/(2π), about 9.5493. The radian is the coherent SI unit of plane angle (BIPM SI Brochure), so rad/s is the SI unit of angular velocity.
Not in an inertial, non-rotating frame of reference: there, the only horizontal force on a cornering car is the inward centripetal force from the tyres. Centrifugal force appears only when motion is described from inside the rotating frame, where it has the same size, mv²/r, pointing outward. The outward push a passenger feels is their body's inertia carrying it in a straight line while the car turns.
I = ½MR² about its central axis, so a 5 kg disc of radius 0.2 m has I = 0.5 × 5 × 0.2² = 0.1 kg·m². Other standard results are MR² for a thin hoop, ⅖MR² for a solid sphere, ⅔MR² for a thin spherical shell, and ML²/12 or ML²/3 for a thin rod about its centre or one end (Serway and Jewett, Table 10.2).
Torque is τ = rF sin θ, where r is the distance from the axis to the point where the force acts and θ is the angle between the arm and the force. A 50 N push at 90° on a 0.3 m wrench gives 15 N·m; the same push along the handle (θ = 0°) gives none. One newton-metre is 0.7376 pound-force feet.
تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 7. مثلًا، يجري التحقق من «1000 kg car at 20 m/s on a 50 m curve» بالرجوع إلى Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4.
OpenStax University Physics Volume 1, §6.3 Centripetal force; §10.5 Calculating moments of inertia; §10.6 Torque; Serway & Jewett, Physics for Scientists and Engineers, Table 10.2 — Moments of inertia of homogeneous rigid objects.
تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 7. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.
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