Q#30
A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s². At what time will the body have kinetic energy same as the initial value if the torque continues to act?Answer:
Angular speed ω = 20 rad/s, Deceleration α = 2 rad/s². Since the torque decelerates rotating body, the body will come to rest but the torque continues, hence the body will again accelerate in opposite direction. When after acceleration it gets an angular speed of -ω, the kinetic energy of the body will again be equal to the initial value because of KE = ½Iω² and it depends on the numerical value of ω.
Now we have Final angular speed, ω' = -ω = -20 rad/s
Initial angular speed = ω = 20 rad/s
Deceleration α = 2 rad/s²
If the time taken be t, from the formula ω'=ω-αt
–ω = ω – αt
αt = 2ω
t = 2ω/α = 2(20/2) = 20 s
Q#31
A light rod of length 1 m is pivoted at its center and two masses of 5 kg and 2 kg are hung from the ends as shown in figure (10-E3). Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.
The moment of inertia of the weights about the pivot,
I = 2(0.5)² + 5(0.5)²
I = 1.75 kg.m²
Net torque about the pivot,
T = 5g(0.5) – 2g(0.5)
= 3g(0.5) = 1.5g N.m
Hence the angular acceleration
α = T/I = 1.5g/1.75 = 6g/7
= 8.4 rad/s²
Q#32
Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.
(a) Find the initial angular acceleration of the rod.
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.
Answer:
(a) Since the mass of the rod is uniformly distributed over its length, it will make no difference in the torque produced about the pivot. Hence the torque T = 1.5g Nm
But the moment of inertia of the system will change. The moment of inertia of the rod will have to be added. The moment of inertia of the rod about the pivot
= ml²/12 = 1(1)²/12 = 1/12 kg.m²
Hence total moment of inertia = 1.75 + 1/12 kg.m²
= 1.75 + 0.083 = 1.833 kg.m²
Hence the initial angular acceleration α = T/I
= 1.5(9.8/1.833) = 8.0 rad/s²
(b) Linear acceleration at the end of the rod,
a = αr = 8.0(0.5) = 4.0 m/s²
Consider the 2.0 kg mass. Weight 2g N is downwards, tension T is upwards and the acceleration a is upwards. Hence,
T – 2g = 2a
T = 2g + 2a = 2(9.8) + 2(4.0) = 19.6 + 8 = 27.6 N
For 5 kg mass, acceleration a is downwrds, hence
5g – T = 5a
T = 5g – 5a
T = 5(9.8) – 5(4) = 29 N
Q#33
Figure (10-E4) shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.
Answer:
Let the accelerations of blocks M and m be a and the tension in the string be T and T' respectively. For the block M, we have,
Mg – T = Ma
T = M(g – a)
And for the block m,
T' = ma
Net torque on the pulley = Tr – T'r = (T – T')r
If the angular acceleration of the pulley be α,
Then, a = αr → α = a/r
(T – T')r = Ia/r
M(g – a) – ma = Ia/r²
Mg = Ma + ma + Ia/r²
a = Mg/(M + m + I/r²)
Q#34
A string is wrapped on a wheel of moment of inertia 0.20 kg-m² and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in figure (10-E5). Find the acceleration of the block.
Answer:
Let the acceleration of the block be a m/s² and the tension in the string be T Newton.
2g – T = 2a
T = 2g – 2a
Torque on the pully
= T(0.10) Nm
= 0.10(2g – 2a)
= 0.20(g – a)
Moment of inertia of the pulley, I = 0.20 kg.m²
The angular acceleration α = Torque/MI
a/r = 0.20(g – a)/0.20 = g – a
a = gr – ar
a + ar = gr
a = gr/(1 + r)
a = 9.8(0.10)/(1 + 0.10)
a = 0.98/1.10
a = 98/110 = 0.89 m/s²
Q#35
Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m². Find the tension in the part of the string joining the pulleys.
Answer:
Let the required tension be T'.
Now the torque on the bigger pulley = T'(0.10 Nm)
Angular acceleration a/r =0.10(T'/0.20) = T'/2
a = rT'/2 =0.10*T'/2 = 0.05T' (i)
Net torque on the smaller pulley
= (T – T')r' = (T – T')(0.05 Nm)
Angular acceleration
a/r' = 0.05(T – T')/I'
0.05T'/0.05 = 0.05(T – T')/0.10 {Value of a from (i)}
T' = 0.5T – 0.5T'
1.5T' = 0.5T
For the weight 2g – T = 2a
T = 2g – 2a = 2g – 2(0.05T') = 2g – 0.10T'
Putting this value in (ii), we get
1.5T' = 0.5(2g – 0.10T')
1.5T' = 9.8 – 0.05T'
1.55T' = 9.8
T' = 9.8/1.55 = 6.32 N
Q#36
The pulleys in figure (10-E6) are identical, each having radius R and moment of inertia I. Find the acceleration of the block M.
Let the acceleration be a.
Angular acceleration of the pulleys = a/R
Let the tension in the string connected to M be T and in the string connected to m be T'. Also the tension in the string between the pulleys be T".
Now Mg – T = Ma
T = M(g – a)
Also, T' – mg = ma
T' = m(g + a)
For pulleys,
α = a/R = (T – T")R/I
a = (T – T")R²/I (i)
Also, a/R = (T" – T)R/I
a = (T" – T')R²/I (ii)
Adding (i) and (ii)
2a = (T – T')R²/I = (Mg – Ma – mg – ma)R²/I
2aI = (Mg – Ma – mg – ma)R² = (M – m)gR² – (M + m)aR²
2aI + (M + m)aR² = (M – m)gR²
a{2I + MR² + mR²} = (M – m)gR²
a = (M – m)gR²/{2I + MR² + mR²}
a = (M – m)g/(M + m + 2I/R²)
{Dividing the numeretor and the denominetor by R²]
α = a/R = (T – T")R/I
a = (T – T")R²/I (i)
Also, a/R = (T" – T)R/I
a = (T" – T')R²/I (ii)
Adding (i) and (ii)
2a = (T – T')R²/I = (Mg – Ma – mg – ma)R²/I
2aI = (Mg – Ma – mg – ma)R² = (M – m)gR² – (M + m)aR²
2aI + (M + m)aR² = (M – m)gR²
a{2I + MR² + mR²} = (M – m)gR²
a = (M – m)gR²/{2I + MR² + mR²}
a = (M – m)g/(M + m + 2I/R²)
{Dividing the numeretor and the denominetor by R²]
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