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.
പുതുക്കിയത് പരിശോധിച്ച ഉദാഹരണങ്ങൾ: 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.
പരിശോധനയുടെ ഉറവിടം: Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4
പരിശോധനയുടെ ഉറവിടം: Python 3.8 decimal: v = √(Fr/m) = √100
പരിശോധനയുടെ ഉറവിടം: Python 3.8 math: ω = 2π·3000/60, v = ωr
പരിശോധനയുടെ ഉറവിടം: 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.
കൃത്യത നിങ്ങളുടെ ഇൻപുട്ടുകളെയും രീതിയുടെ അനുമാനങ്ങളെയും ആശ്രയിച്ചിരിക്കുന്നു. ദശാംശ ഗണിതം 50 സാർഥക അക്കങ്ങൾ ഉപയോഗിക്കുന്നു. എന്നാൽ അനുമാനക്കണക്കുകൾ, സംഖ്യാത്മക രീതികൾ, ഉറവിട ഡാറ്റ എന്നിവയ്ക്ക് കൃത്യത കുറവാകാം; പ്രദർശിപ്പിക്കുന്ന മൂല്യം റൗണ്ട് ചെയ്യുന്നത് ഈ പരിമിതികൾ നീക്കില്ല. സ്വതന്ത്ര ഉറവിടങ്ങളിലെ പരിഹാരങ്ങളുമായി പരിശോധിച്ച ഉദാഹരണങ്ങൾ: 7. ഉദാഹരണത്തിന്, “1000 kg car at 20 m/s on a 50 m curve” എന്നത് Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4 ഉപയോഗിച്ച് പരിശോധിക്കുന്നു.
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.
ഈ കാൽക്കുലേറ്ററിൽ സ്വതന്ത്ര ആധാരങ്ങളിൽ നിന്ന് ഉത്തരങ്ങൾ ലഭിച്ച 7 പരിഹരിച്ച ഉദാഹരണങ്ങളുണ്ട്. ഇവ പരിശോധനാസമുച്ചയത്തിൽ പ്രവർത്തിക്കുന്നു; നിങ്ങൾക്ക് ഇവിടെയും പ്രവർത്തിപ്പിക്കാം.
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