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3. ROTATIONAL MOTION 1. A circular disc of mass 10 kg and radius 0.2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. Calculate the angular velocity of the disc at the end of 6 s from the rest. Given : τ = 10 Nm M = 10 kg R = 0.2 m, t = 6s = 0 ω1 To Find : ω2 = ? Formula : i) I = ii) τ = iii) ω2 = Solution : Since M.I of disc, I ∴ = I = I Now, τ = = ∴ α = ∴ α ω2 ω2 ω2 = = = = 2. 1 MR2 2 Iα ω1 + αt 1 MR2 2 Given : D = ∴ = M ω1 ω2 τ To Find : i) ii) Formula : i) ∴ A solid sphere of diameter 25cm and mass 25 kg rotates about an axis through its centre. Calculate its moment of inertia, if its angular velocity changes from 2 rad/s to 12 rad/s in 5 second. Also calculate the torque applied. = = = = I τ = = = 0.25 m ? ? M.I of sphere, I = 2 MR2 5 Iα 2 MR2 5 I = I = I = 2 0.25 × 25 × 5 2 0.1562 kg m2 τ = Iα τ = ω2 – ω1 I t = τ 3. = 25 cm 2 25 kg 2 rad/s 12 rad/s 5s ii) τ Solution : 1 × 10 × (0.2)2 2 0.2 kg m2 Iα τ 10 = I 0.2 50 rad/s ω1 + αt 0 + 50 × 6 300 rad/sec R 25 cm = 2 ω1 – ω2 ∵ α = t 12 – 2 0.1562 × 5 0.3124 Nm Calculate moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation coinciding with its diameter and tangent perpendicular to its plane. Given : M = 500 g = 0.5 kg R = 0.5 m Rotational Motion MAHESH TUTORIALS SCIENCE .. 14 ? ? i) Id = ii) IT Solution : = ∴ ∴ 4. Id = Id = Id IT IT IT = = = = = MR 2 2 2MR2 2 0.0625 kg m2 6.25 × 10–2 kg m2 2MR2 2 × 0.5 × (0.5)2 0.25 kg m2 2 R. Show that the 5 radius of gyration about a tangential its diameter is axis of rotation is = = Kr = ∴ Kd = Id + MR2 ∴ MKT2 = MK2d + MR2 ∵ I T = MK 2T I = MK 2 d d KT2 = K2d + R2 2 T K = 2 R + R2 5 KT2 = 2 R2 + R2 5 ∴ KT = 7 2 R 5 ∴ KT = ∴ 2 2 A solid sphere has a radius ‘R’. If the radius of gyration of this sphere about Given : M Kd = MR 2 2 0.5 × ( 0.5 ) 7 R. 5 mass of uniform solid sphere Radius of gyration about diameter Radius of gyration about tangent 2 R 5 = A ballet dancer spins about a vertical axis at 90 r.p.m with arms outstretched. With the arms folded, the moment of inertia about the same axis of rotation changes to 75 %. Calculate the new speed of rotation. Given : n1 = 90 r.p.m, I1 = M.I with arms out stretched = 0.75I1 I2 To Find : n2 = ? Formula : I 1 ω1 = I 2 ω2 Solution : I1 (2π π n1) = I2 (2π π n2) [... ω = 2π πn] ∴ n2 7 R 5 Proof : From the theorem of parallel axes, IO = I c + Mh2 Rotational Motion 7 R 5 5. To Prove : KT [ . . . h = R] IT ∴ To Find : Id = IT = Formula : = I1 n1 I2 = I1 × 90 0.75I 1 = ∴ n2 = 100 × 90 75 120 r.p.m MAHESH TUTORIALS SCIENCE 6. .. 15 A torque of 400 Nm acting on a body Solution : of mass 40 kg produces an angular 2 acceleration of 20 rad/s2. Calculate the I1 = MR21 5 moment of inertia and radius of gyration of the body. 2 I2 = MR22 Given : 5 τ = 400 Nm 2 2 M = 40 kg R1 R1 I1 2 ∴ = = α = 20 rad/s I2 R2 2R 1 To Find : I = ? 2 1 1 K = ? = = 4 2 Formula : i) τ = Iα I1 ∴ = 1:4 I I2 ii) K = M 8. A flywheel in the form of a disc, rotating Solution : about an axis passing through its centre τ 400 and perpendicular to its plane, looses I = = α 20 100 J of energy, when slowing down 2 from 60 r.p.m to 30 r.p.m. Find its I = 20 kg m moment of inertia about the same axis I and change in its angular momentum Now, K = M Given : n1 = 60 r.p.m. 20 1 ∴ K = = 60 40 2 = = 1 r.p.s, 60 ∴ K = 0.5 n2 = 30 r.p.m. ∴ K = 0.707 m 30 1 = = r.p.s, 60 2 7. If the radius of solid sphere is doubled ∆E = 100 J by keeping its mass constant, compare To Find : the moment of inertia about any i) I = ? diameter. ii) ∆L = ? Given : Formula : R2 = 2R1 1 M = constant i) K.E. = Iω2 To Find : 2 ii) L = Iω I1 = ? Solution : I2 1 Formula : i) KE1 = × I × (2πn1)2 2 2 I = MR2 = 2π2n21I 5 Similarly, K.E2 = 2π2n22I Rotational Motion MAHESH TUTORIALS SCIENCE .. 16 ∴ ∆E ∴ ∆E = = = ∴ I = = = = ∴ ∴ ∴ 9. . E 2 2 2 2 2 2 2 ∆E 2 π n 22 – n 12 2 100 2 2 1 2 ( 3.14 ) – 12 2 ∆L = = Formula : L = Iω = constant Solution : L′ = L′ ′ ′ ′ I ω = I ′′ ω ′ ′ I′(ω2 – ω2 ) = (I + I) ω′′ I × 2π (n2 – n1) = 2I × 2πn′′ n2 – n1 = 2n ′′ 240 – 120 2 n′ ′ –100 23 2 ( 3.14 ) 4 –200 3 ( 3.142 ) 2 I = – 6.753 kg m2 Negative sign shows that energy is lost I = 6.759 kg m2 L = Iω L = I (2πn) (∵ ω = 2πn) L = 2πIn ∆L = L2 – L1 = 2π I (n2 – n1) = ∴ – K.E1 (2π n – 2π2n21) I 2π (n – n21) I K 1 2 × 3.14 × 6.753 – 1 2 – 3.142 × 6.753 21.21 kg m2/s = n′ ′ = 60 rpm 10. A uniform circular disc with its plane horizontal is rotating about a vertical axis passing through its centre at a speed of 180 r.p.m. A small piece of wax of mass 1.9 g falls vertically on the disc and sticks to it at a distance of 25 cm from the axis. If the speed of rotation is now reduced by 60 r.p.m., calculate moment of inertia of the disc. Given : n1 = 180 r.p.m = n2 = = m h Two wheels of moment of inertia 4 kg m2 rotate side by side at the rate of I1 120 rev/min and 240 rev/min and respectively in the opposite directions. I2 If now both the wheels are coupled by means of weightless shaft so that both To find : the wheels now rotate with a common I1 angular speed. Find the new speed of Formula : rotation. I1ω1 Given : 2 I = 4 kg m I2 n1 = 120 rpm n2 = 240 rpm Solution : To find : I2 n′ ′ = ? Rotational Motion = = = = = 180 = 3 r.p.s, 60 (180 – 60) r.p.m 120 = 2 r.p.s, 60 1.9 g = 1.9 × 10–3 kg, r 25 cm = 0.25 m M.I of the disc about vertical axis M.I of disc about same axis with wax = ? = I 2 ω2 = I1 + Iwax (Parallel axes theorem) = I1 + Iwax MAHESH TUTORIALS SCIENCE ∴ ∴ ∴ I2 Consider, I 1 ω1 I1 (2πn1) I1n1 I1 (n1 – n2) .. 17 I1 + mr2 = I 2 ω2 (I1 + mr2) × (2πn2) (I1 + mr2) n2 mr2n2 = = = = Ml 2 Ml 2 + 12 16 = 7 7 Ml2 = (1) (1)2 48 48 0.1458 kg m2 = ∴ Io = K = K = 2 ∴ I1 = mr n 2 n1 – n2 2 = I1 = 1.9 × 10 – 3 × ( 0.25 ) × 2 3–2 2.375 × 10–4 kg m2 11. A thin uniform rod of length 1 m and mass 1 kg is rotating about an axis passing through its centre and perpendicular to its length. Calculate moment of inertia and radius of gyration of the rod about an axis passing through a point mid way between the centre and its edge, perpendicular to its length. Given : M = 1 kg l = 1m h = To find : i) I ii) K Formula : i) I0 ii) K ∴ ? ? = Ic + Mh2 = A homogenous rod XY of length L and mass ‘M’ is pivoted at the centre ‘C’ such that it can rotate freely in vertical plane. Intially the rod is in the horizontal position. A blob of wax of same mass ‘M’ that of the rod falls vertically with the speed ‘V’ and sticks to the rod midway between points C and Y. If the rod rotates with angular speed ‘ω’ what will be angular speed in terms of V and L ? Solution : ITotal = Irod + Iwax ∴ I M Solution : M. I of rod about an axis through its ∴ centre is given by ∴ 0.1458 1 12. l 4 = = I = M 0.3818 m = ML2 + Mr2 12 = ML2 ML2 + 12 4 L ∵ r = 2 7ML2 48 Since angular momentum is conserved, Initial angular = Final angular momentum momentum ITotal = L MV = 4 7ML2 48 Ic = Ml 2 12 ∴ V 4 = 7L ω 48 I0 = Ic + Mh2 ∴ ω = V 48 × 4 7L I0 = l Ml 2 +M 12 4 ∴ ω = 12 V 7 L 2 ω Rotational Motion