The bob of a pendulum at rest is given a horizontal velocity of √ 7gl􏰌 , where 􏰌 is l length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob)

Question

The bob of a pendulum at rest is given a horizontal velocity of √ 7gl􏰌 , where 􏰌 is l length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob) (A) 2 mg (B) mg (C) 4 mg (D) 7 mg

Tardigrade Admin · Accepted Answer

From energy conservation (1/2)mv21+mg.2l=(1/2)mv2 (1/2)mv21=(7/2)mgl-2mgl mv21=3mgl...(1) Now, T = (mv21/l)-mg=(3mgl/l)-mg T = 2mg

Q. The bob of a pendulum at rest is given a horizontal velocity of $7 g l$ 􏰌 , where 􏰌 is $l$ length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob)

2 mg

mg

4 mg

7 mg

From energy conservation
$\frac{1}{2} m v_{1}^{2} + m g .2 l = \frac{1}{2} m v^{2}$
$\frac{1}{2} m v_{1}^{2} = \frac{7}{2} m g l - 2 m g l$
$m v_{1}^{2} = 3 m g l ... (1)$
Now,
T = $\frac{m v _{1}^{2}}{l} - m g = \frac{3 m g l}{l} - m g$
T = 2mg

Q. The bob of a pendulum at rest is given a horizontal velocity of 7gl​􏰌 , where 􏰌 is l length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob)

2 mg

mg

4 mg

7 mg

From energy conservation 21​mv12​+mg.2l=21​mv2 21​mv12​=27​mgl−2mgl mv12​=3mgl...(1) Now, T = lmv12​​−mg=l3mgl​−mg T = 2mg

Q. The bob of a pendulum at rest is given a horizontal velocity of $7 g l$ 􏰌 , where 􏰌 is $l$ length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob)

From energy conservation
$\frac{1}{2} m v_{1}^{2} + m g .2 l = \frac{1}{2} m v^{2}$
$\frac{1}{2} m v_{1}^{2} = \frac{7}{2} m g l - 2 m g l$
$m v_{1}^{2} = 3 m g l ... (1)$
Now,
T = $\frac{m v _{1}^{2}}{l} - m g = \frac{3 m g l}{l} - m g$
T = 2mg