A pendulum consisting of a massless
string of length 20 cm and a tiny bob
of mass 100 g is set up as a conical
pendulum. Its bob now performs 75 rpm.
Calculate kinetic energy and increase in
the gravitational potential energy of the
bob. (Use a? = 10)
Answers
Answer:
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Answer:
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,g = 10 m/s2, π2 = 10,
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,g = 10 m/s2, π2 = 10,T = 1n=45s=0.8s
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,g = 10 m/s2, π2 = 10,T = 1n=45s=0.8sT = 2πL cosθg
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,g = 10 m/s2, π2 = 10,T = 1n=45s=0.8sT = 2πL cosθg∴ T2 = 4πL cosθg
Data: L = 0.2 m, m = 0.1 kg, n = 7560=54rps,g = 10 m/s2, π2 = 10,T = 1n=45s=0.8sT = 2πL cosθg∴ T2 = 4πL cosθg∴ h = L cos θ = gT24π2
=(10)(0.8)24(10)=0.16 m ...(1)
...(1)∴ cos θ = 0.160.2
=0.8
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)= 0.045 J
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)= 0.045 JThe increase in gravitational PE,
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)= 0.045 JThe increase in gravitational PE,Δ PE = mg (L - h)
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)= 0.045 JThe increase in gravitational PE,Δ PE = mg (L - h)= (0.1)(10)(0.2 - 0.16)
=0.8∴ θ = cos-1 0.8 = 36.87° = 36°5'v2 = rg tan θ = (L sin θ)(g) tan 36.87°= (0.12)(10)(0.7500)= 0.9The KE of the bob =12mv2=12(0.1)(0.9)= 0.045 JThe increase in gravitational PE,Δ PE = mg (L - h)= (0.1)(10)(0.2 - 0.16)= 0.04 J